Chapter 3 – Polynomials
Practice set 3.1
1. State whether the given algebraic expressions are polynomials? Justify.
(i) y + \(\large \frac {1}{y}\)
Solution:
y + \(\large \frac {1}{y}\)
= y + y⁻¹
Here one of the power of y is – 1, which is not a whole number.
∴ It is not a polynomial
(ii) 2 – 5 \(\sqrt{x}\)
Solution:
2 – 5 \(\sqrt{x}\)
= 2 – 5 \(x^{\frac{1}{2}\)
Here the power of x is \(\large \frac {1}{2}\), which is not a whole number.
∴ It is not a polynomial.
(iii) x² + 7x + 9
Solution:
x² + 7x + 9
Here all powers of x are whole numbers.
∴ It is a polynomial.
(iv) 2m⁻² + 7m – 5
Solution:
2m⁻² + 7m – 5
Here one of the power of m is –2 which is not a whole number.
∴ It is not a polynomial.
(v) 10
Solution:
10 = 10 × x⁰
Here the power of x is a whole number.
∴ It is a polynomial.
2. Write the coefficient of m³ in each of the given polynomial.
(i) m³
Ans: Coefficient of m³ : 1
(ii) – \(\large \frac {3}{2}\) + m – \(\sqrt{3}\) m³
Ans: Coefficient of m³ : – \(\sqrt{3}\)
(iii) – \(\large \frac {2}{3}\) m³ – 5m² + 7m – 1
Ans: Coefficient of m³ : – \(\large \frac {2}{3}\)
3. Write the polynomial in x using the given information.
(i) Monomial with degree 7
Ans: 51x⁷
(ii) Binomial with degree 35
Ans: 2x³⁵ + 36
(iii) Trinomial with degree 8
Ans: x⁸ – 2x⁷ + 3x⁶
4. Write the degree of the given polynomials.
(i) 5
Ans: Degree of Polynomial : 0
(ii) x°
Ans: Degree of Polynomial : 0
(iii) x²
Ans: Degree of Polynomial : 2
(iv) \(\sqrt{2}\) m¹⁰ – 7
Ans: Degree of Polynomial : 10
(v) 2p – 7
Ans: Degree of Polynomial : 1
(vi) 7y – y³ + y⁵
Ans: Degree of Polynomial : 5
(vii) xyz + xy – z
Ans: Degree of Polynomial : 3
(viii) m³n⁷ – 3m⁵n + mn
Ans: Degree of Polynomial : 10
5. Classify the following polynomials as linear, quadratic and cubic polynomial.
(i) 2x² + 3x + 1
Ans: Quadratic Polynomial
(ii) 5p
Ans: Linear Polynomial
(iii) \(\sqrt{2}\) y – 12
Ans: Linear Polynomial
(iv) m³ + 7m² + \(\large \frac {5}{2}\) m – \(\sqrt{7}\)
Ans: Cubic Polynomial
(v) a²
Ans: Quadratic Polynomial
(vi) 3r³
Ans: Cubic Polynomial
6. Write the following polynomials in standard form.
(i) m³ + 3 + 5m
Ans: The standard form is:
m³ + 5m + 3
(ii) – 7y + y⁵ + 3y³ – \(\large \frac {1}{2}\) + 2y⁴ – y²
Ans: The standard form is:
y⁵ + 2y⁴ + 3y³ – y² – 7y – \(\large \frac {1}{2}\)
7. Write the following polynomials in coefficient form.
(i) x³ – 2
Solution:
The index form of the given polynomial is
x³ + 0x² + 0x – 2
∴ The coefficient form of the given polynomial is (1, 0, 0, – 2)
(ii) 5y
Solution:
The index form of the given polynomial is 5y + 0
∴ The coefficient form of the given polynomial is (5, 0)
(iii) 2m⁴ – 3m² + 7
Solution:
The index form of the given polynomial is
2m⁴ + 0m³ – 3m² + 0m + 7
∴ The coefficient form of the given polynomial is (2, 0, –3, 0, 7)
(iv) – \(\large \frac {2}{3}\)
Solution:
The coefficient form of the given polynomial is – \(\large \frac {2}{3}\)
8. Write the polynomials in standard form.
(i) (1, 2, 3)
Solution:
The polynomial (1, 2, 3) contains 3 coefficients.
∴ The degree of the polynomial is 3 – 1 = 2
∴ The index form of the given polynomial is x² + 2x + 3
(ii) (5, 0, 0, 0, – 1)
Solution:
The polynomial (5, 0, 0, 0, –1) contains 5
coefficients.
∴ The degree of the polynomial is 5 – 1 = 4
∴ The index form of the given polynomial is 5x⁴ + 0x³ + 0x² + 0x – 1 = 5x⁴ – 1
(iii) (– 2, 2, – 2, 2)
Solution:
The polynomial (–2, 2, –2, 2) contains 4 coefficients.
∴ The degree of the polynomial is 4 – 1 = 3
∴ The index form of the given polynomial is – 2x³ + 2x² – 2x + 2
9. Write the appropriate polynomials in the boxes.
Solution:
Practice set 3.2
(1) Use the given letters to write the answer.
(i) There are ‘a’ trees in the village Lat. If the number of trees increases every year by ‘b’, then how many trees will there be after ‘x’ years?
Solution:
Number of trees in a village = a
Yearly increase in number of trees = b
∴ Increase in number of trees after x years = bx
∴ Number of trees in a village after x years = a + bx
(ii) For the parade there are y students in each row and x such row are formed. Then, how many students are there for the parade in all ?
Solution:
Number of rows = x
Number of students in each row = y
∴ Total number of students participated in parade
= Number of rows × Number of students in each row
= x × y
= xy
(iii) The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.
Solution:
Digit at Ten’s place = m
Digit at Unit’s place = n
∴ Two digit number = 10m + n
(2) Add the given polynomials.
(i) x³ – 2x² – 9 ; 5x³ + 2x + 9
Solution:
x³ – 2x² – 9 + 5x³ + 2x + 9
= x³ + 5x³ – 2x² + 2x – 9 + 9
= 6x³ – 2x² + 2x
(ii) – 7m⁴ + 5m³ + \(\sqrt{2}\) ; 5m⁴ – 3m³ + 2m² + 3m – 6
Solution:
– 7m⁴ + 5m³ + \(\sqrt{2}\) + 5m⁴ – 3m³ + 2m² + 3m – 6
= – 7m⁴ + 5m⁴ + 5m³ – 3m³ + 2m² + 3m + \(\sqrt{2}\) – 6
= – 2m⁴ + 2m³ + 2m² + 3m + \(\sqrt{2}\) – 6
(iii) 2y² + 7y + 5 ; 3y + 9 ; 3y² – 4y – 3
Solution:
2y² + 7y + 5 + 3y + 9 + 3y² – 4y – 3
= 2y² + 3y² + 7y + 3y – 4y + 5 + 9 – 3
= 5y² + 6y + 11
(3) Subtract the second polynomial from the first.
(i) x² – 9x + \(\sqrt{3}\) ; – 19x + \(\sqrt{3}\) + 7x²
Solution:
x² – 9x + \(\sqrt{3}\) – (– 19x + \(\sqrt{3}\) + 7x²)
= x² – 9x + \(\sqrt{3}\) + 19x – \(\sqrt{3}\) – 7x²
= x² – 7x² – 9x + 19x + \(\sqrt{3}\) – \(\sqrt{3}\)
= – 6x² + 10x
(ii) 2ab² + 3a²b – 4ab ; 3ab – 8ab² + 2a²b
Solution:
2ab² + 3a²b – 4ab –(3ab – 8ab² + 2a²b)
= 2ab²+ 3a²b – 4ab – 3ab + 8ab² – 2a²b
= 2ab² + 8ab² + 3a²b – 2a²b – 4ab – 3ab
= 10ab² + a²b – 7ab
(4) Multiply the given polynomials.
(i) 2x ; x² – 2x – 1
Solution:
2x × (x² – 2x – 1)
= 2x³ – 4x² – 2x
(ii) x⁵ – 1 ; x³ + 2x² + 2
Solution:
(x⁵ – 1) × (x³ + 2x² + 2)
= x⁵ (x³ + 2x² + 2) – 1 (x³ + 2x² + 2)
= x⁸ + 2x⁷ + 2x⁵ – x³ – 2x² – 2
(iii) 2y + 1; y² – 2y³ + 3y
Solution:
(2y + 1) × (y² – 2y³ + 3y)
= 2y (y² – 2y³ + 3y) + 1 (y² – 2y³ + 3y)
= 2y³ – 4y⁴ + 6y² + y² – 2y³ + 3y
= – 4y⁴ + 7y² + 3y
(5) Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor × Quotient + Remainder’.
(i) x³ – 64; x – 4
Solution:
(x³ – 64) ÷ (x – 4) = x² + 4x + 16
∴ x³ – 64 = (x – 4) × (x² + 4x + 16) + 0
(ii) 5x⁵ + 4x⁴ – 3x³ + 2x² + 2; x² – x
Solution:
(5x⁵ + 4x⁴ – 3x³ + 2x² + 2) ÷ (x² – x) = (x² – x) (5x³ + 9x² + 6x+ 8) + (8x + 2)
∴ 5x⁵ + 4x⁴ – 3x³ + 2x² + 2 = (x² – x) (5x³ + 9x² + 6x+ 8) + (8x + 2)
(6*) Write down the information in the form of algebraic expression and simplify.
There is a rectangular farm with length (2a² + 3b²) metre and breadth (a² + b²) metre. The farmer used a square shaped plot of the farm to build a house. The side of the plot was (a² – b²) metre. What is the area of the remaining part of the farm ?
Solution:
For rectangular field,
Length = (2a² + 3b²) m
Breadth = (a + b²) m
Area of rectangular field
= Length × Breadth
= (2a² + 3b²)(a² + b²)
= 2a² × (a² + b²) + 3b² × (a² + b²)
= 2a⁴ + 2a² × b² + 3a² × b² + 3b⁴
= (2a⁴ + 5a² × b² + 3b⁴) sq.m
For square house,
Side = (a² – b²) m
∴ Area of square house
= (side)²
= (a² – b²)²
= (a²)² – 2a² × b² + (b²)²
= (a⁴ – 2a² × b² + b⁴) × sq .m
Area of remaining field
= Area of rectangular field – Area of square house
= 2a + 5a²b² + 3b⁴ – (a – 2a²b² + b⁴)
= 2a + 5a²b² + 3b – a² + 2ab² – b
= 2a – a + 5ab² + 2a²b² + 3b – b
= a + 7a²b² + 2b
∴ Area of remaining field is (a + 7a²b² + 2b) sq.m.
Practice set 3.3
1. Divide each of the following polynomials by synthetic division method and also by linear division method. Write the quotient and the remainder.
(i) (2m² – 3m + 10) ÷ (m – 5)
Solution:
Dividend = 2m² – 3m + 10
Dividend in coefficient form = (2, – 3, 10)
Comparing divisor m – 5 with m – a, we get, a = 5
Quotient in coefficient form is (2, 7)
∴ Quotient = 2m + 7, Remainder = 45
Linear method:
2m² – 3m + 10
= 2m(m – 5) + 10m – 3m + 10
= 2m(m – 5) + 7m + 10
= 2m(m – 5) + 7(m – 5) + 35 + 10
= 2m(m – 5) + 7(m – 5) + 45
∴ 2m² – 3m + 10 = (m – 5) (2m + 7) + 45
Comparing with Dividend = Divisor × Quotient + Remainder, we get,
∴ Quotient = 2m + 7, Remainder = 45
(ii) (x⁴ + 2x³ + 3x² + 4x + 5) ÷ (x + 2)
Solution:
Dividend = x⁴ + 2x³ + 3x² + 4x + 5
Dividend in coefficient form = (1, 2, 3, 4, 5)
Comparing divisor x + 2 with x – a, we get, a = – 2
Quotient in coefficient form is (1, 0, 3, –2)
∴ Quotient = x³ + 0x² + 3x – 2 = x³ + 3x – 2, Remainder = 9
Linear method:
x⁴ + 2x³ + 3x² + 4x + 5
= x³(x + 2) – 2x³ + 2x³ + 3x² + 4x + 5
= x³(x + 2) + 3x² + 4x + 5
= x³(x + 2) + 3x(x + 2) – 6x + 4x + 5
= x³(x + 2) + 3x(x + 2) – 2x + 5
= x³(x + 2) + 3x(x + 2) – 2(x + 2) + 4 + 5
= x³(x + 2) + 3x(x + 2) – 2(x + 2) + 9
∴ x⁴ + 2x³ + 3x² + 4x + 5 = (x + 2) (x³ + 3x – 2) + 9
Comparing with Dividend = Divisor × Quotient + Remainder, we get,
∴ Quotient = x³ + 3x – 2, Remainder = 9
(iii) (y³ – 216) ÷ (y – 6)
Solution:
Dividend = y³ – 216
Dividend in index form = y³ + 0y² + 0y – 216
∴ Dividend in coefficient form = (1, 0, 0, – 216)
Comparing divisor y – 6 with y – a, we get, a = 6
Quotient in coefficient form is (1, 6, 36)
∴ Quotient = y² + 6y + 36, Remainder = 0.
Linear method:
y³ – 216
= y²(y – 6) + 6y² – 216
= y²(y – 6) + 6y(y – 6) + 36y – 216
= y²(y – 6) + 6y(y – 6) + 36(y – 6) + 216 – 216
∴ y³ – 216 = (y – 6)(y² + 6y + 36) – 0
Comparing with Dividend = Divisor × Quotient + Remainder, we get,
Quotient = y² + 6y + 36, Remainder = 0.
(iv) (2x⁴ + 3x³ + 4x – 2x²) ÷ (x + 3)
Solution:
Dividend = 2x⁴ + 3x³ + 4x – 2x²
Dividend in index form = 2x⁴ + 3x³ – 2x² + 4x + 0
∴ Dividend in coefficient form = (2, 3, –2, 4, 0)
Comparing divisor x + 3 with x – a, we get, a = – 3
Quotient in coefficient form is (2, – 3, 7, –17)
∴ Quotient = 2x³ – 3x² + 7x – 17, Remainder = 51
Linear method:
2x⁴ + 3x³ + 4x – 2x²
= 2x⁴ + 3x³ – 2x² + 4x
= 2x³(x + 3) – 6x³ + 3x³ – 2x² + 4x
= 2x³(x + 3) – 3x³ – 2x² + 4x
= 2x³(x + 3) – 3x²(x + 3) + 9x² – 2x² + 4x
= 2x³(x + 3) – 3x²(x + 3) + 7x² + 4x
= 2x³(x + 3) – 3x² (x + 3) + 7x (x + 3) – 21x + 4x
= 2x³(x + 3) – 3x² (x + 3) + 7x(x + 3) – 17x
= 2x³(x + 3) – 3x²(x + 3) + 7x (x + 3) – 17(x + 3) + 51
∴ 2x⁴ + 3x³ + 4x – 2x² = (x + 3) (2x³ – 3x² + 7x – 17) + 51
Comparing with Dividend = Divisor × Quotient + Remainder, we get,
∴ Quotient = 2x³ – 3x² + 7x – 17, Remainder = 51
(v) (x⁴ – 3x² – 8) ÷ (x + 4)
Solution:
Dividend = x⁴ – 3x² – 8
Dividend in index form = x⁴ + 0x³ – 3x² + 0x – 8
∴ Dividend in coefficient form = (1, 0, – 3, 0, – 8)
Comparing divisor x + 4 with x – a, we get, a = – 4
Quotient in coefficient form is (1, – 4, 13, – 52)
∴ Quotient = x³– 4x² + 13x – 52, Remainder = 200
Linear method:
x⁴ – 3x² – 8
= x³(x + 4) – 4x³ – 3x² – 8
= x³(x + 4) – 4x²(x + 4) + 16x² – 3x² – 8
= x³(x + 4) – 4x²(x + 4) + 13x² – 8
= x³(x + 4) – 4x²(x + 4) + 13x(x + 4) – 52x – 8
= x³(x + 4) – 4x²(x + 4) + 13x(x + 4) – 52 (x + 4) + 208 – 8
∴ x⁴ – 3x² – 8 = (x + 4) (x³ – 4x² + 13x – 52) + 200
Comparing with Dividend = Divisor × Quotient + Remainder, we get,
∴ Quotient = x³ – 4x²+ 13x – 52, Remainder = 200
(vi) (y³ – 3y² + 5y – 1) ÷ (y – 1)
Solution:
Dividend = y³ – 3y² + 5y – 1
∴ Dividend in coefficient form = (1, – 3, 5, – 1)
Comparing divisor y – 1 with y – a, we get, a = 1
Quotient in coefficient form is (1, – 2, 3)
∴ Quotient = y² – 2y + 3, Remainder = 2
Linear method:
y³ – 3y² + 5y – 1
= y²(y – 1) + y² – 3y² + 5y – 1
= y²(y – 1) – 2y² + 5y – 1
= y²(y – 1) – 2y(y – 1) – 2y + 5y – 1
= y²(y – 1) – 2y(y – 1) + 3y – 1
= y²(y – 1) – 2y(y – 1)+ 3(y – 1) + 3 – 1
∴ y³ – 3y² + 5y – 1 = (y – 1)(y² – 2y + 3) + 2
Comparing with Dividend = Divisor × Quotient + Remainder, we get,
∴ Quotient = (y² – 2y + 3), Remainder = 2
Practice set 3.4
(1) For x = 0 find the value of the polynomial x² – 5x + 5.
Solution:
p(x) = x² – 5x + 5
∴ p(0) = 0² – 5 × 0 + 5
∴ p(0) = 5
Ans: The value of given polynomial when x = 0 is 5.
(2) If p(y) = y² – 3 \(\sqrt{2}\) y + 1 then find p (3 \(\sqrt{2}\)).
Solution:
p(y) = y² – 3 \(\sqrt{2}\) y + 1
∴ p(3 \(\sqrt{2}\)) = (3 \(\sqrt{2}\))² – 3 \(\sqrt{2}\) × 3 \(\sqrt{2}\) + 1
∴ p(3 \(\sqrt{2}\)) = 18 – 18 + 1
∴ p(3 \(\sqrt{2}\)) = 1
Ans: The value of given polynomial when
y = 3 \(\sqrt{2}\) is 1.
(3) If p(m) = m³ + 2m² – m + 10 then p(a) + p(– a) = ?
Solution:
p(m) = m³ + 2m² – m + 10
∴ p(a) = a³ + 2a² – a + 10
and p(– a) = (– a)³ + 2(– a)² – (– a) + 10
∴ p(– a) = – a³ + 2a² + a + 10
∴ p(a) + p(–a) = a³ + 2a² – a +10 – a³ + 2 a² + a + 10
∴ p(a) + p(– a) = 4a² + 20
Ans: The value of p(a) + p(– a) is 4a² + 20.
(4) If p(y) = 2y³ – 6y² – 5y + 7 then find p(2).
Solution:
p(y) = 2y³ – 6y² – 5y + 7
∴ p(2) = 2(2)³ – 6(2)² – 5(2) + 7
∴ p(2) = 16 – 24 – 10 + 7
∴ p(2) = – 11
Ans: The value of p(2) is – 11.
Practice set 3.5
(1) Find the value of the polynomial 2x – 2x³ + 7 using given values for x.
(i) x = 3
Solution:
p(x) = 2x – 2x³ + 7
∴ p(3) = 2(3) – 2(3)³ + 7
∴ p(3) = 6 – 54 + 7
∴ p(3) = – 48 + 7
∴ p(3) = – 41
(ii) x = – 1
Solution:
p(x) = 2x – 2x³ + 7
∴ p(– 1) = 2(– 1) – 2(– 1)³ + 7
∴ p(– 1) = – 2 + 2 + 7
∴ p(– 1) = 7
(iii) x = 0
Solution:
p(x) = 2x – 2x³ + 7
∴ p(0) = 2(0) – 2(0)³ + 7
∴ p(0) = 0 – 0 + 7
∴ p(0) = 7
(2) For each of the following polynomial, find p(1), p(0) and p(– 2).
(i) p(x) = x³
Solution:
p(1) = 1³
∴ p(1) = 1
p(0) = 0³
∴ p(0) = 0
p(– 2) = (– 2)³
∴ p(– 2) = – 8
(ii) p(y) = y² – 2y + 5
Solution:
p(1) = (1)² – 2(1) + 5
∴ p(1) = 1 – 2 + 5
∴ p(1) = 4
p(0) = (0)² – 2(0) + 5
∴ p(0) = 0 – 0 + 5
∴ p(0) = 5
p(– 2) = (– 2)² – 2(– 2) + 5
∴ p(– 2) = 4 + 4 + 5
∴ p(– 2) = 13
(iii) p(x) = x⁴ – 2x² – x
Solution:
p(1) = (1)⁴ – 2(1)² – (1)
∴ p(1) = 1 – 2 – 1
∴ p(1) = – 2
p(0) = (0)⁴ – 2(0)² – (0)
∴ p(0) = 0 – 0 – 0
∴ p(0) = 0
p(– 2) = (– 2)⁴ – 2(– 2)² – (– 2)
∴ p(– 2) = 16 – 8 + 2
∴ p(– 2) = 10
(3) If the value of the polynomial m³ + 2m + a is 12 for m = 2, then find the value of a.
Solution:
p(m) = m³ + 2m + a
p(2) = (2)³ + 2(2) + a
∴ p(2) = 8 + 4 + a
∴ p(2) = 12 + a
But p(2) = 12 …[Given]
∴ 12 + a = 12
∴ a = 12 – 12
∴ a = 0
Ans: The value of a is 0
(4) For the polynomial mx² – 2x + 3 if p(– 1) = 7 then find m.
Solution:
p(x) = mx² – 2x + 3
∴ p(–1) = m(–1)² – 2(–1) + 3
∴ p(– 1) = m + 2 + 3
∴ p(– 1) = m + 5
But p(– 1) = 7 …[Given]
∴ m + 5 = 7
∴ m = 7 – 5
∴ m = 2
Ans: The value of m is 2
(5) Divide the first polynomial by the second polynomial and find the remainder using remainder theorem.
(i) (x² – 7x + 9) ; (x + 1)
Solution:
p(x) = x² – 7x + 9
Divisor = x + 1
Put x = – 1 in p(x)
∴ p(– 1) = (– 1)² – 7(– 1) + 9
∴ p(– 1) = 1 + 7 + 9
∴ p(– 1) = 17
By remainder theorem, p(– 1) = remainder
∴ Remainder = 17
(ii) (2x³ – 2x² + ax – a) ; (x – a)
Solution:
p(x) = 2x³ – 2x² + ax – a
Divisor = x – a
Put x = a in p(x)
∴ p(a) = 2(a)³ – 2(a)² + a(a) – a
∴ p(a) = 2a³ – 2a² + a² – a
∴ p(a) = 2a³ – a² – a
By remainder theorem, p(a) = remainder
∴ Remainder = 2a³ – a² – a
(iii) (54m³ + 18m² – 27m + 5) ; (m – 3)
Solution:
p(m) = 54m³ + 18m² – 27m + 5
Divisor = m – 3
Put m = 3 in p(m)
∴ p(m) = 54(3)³ + 18(3)² – 27(3) + 5
∴ p(m) = 1458 + 162 – 81 + 5
∴ p(m) = 1544
By remainder theorem, p(m) = remainder
∴ Remainder = 1544
(6) If the polynomial y³ – 5y² + 7y + m is divided by y + 2 and the remainder is 50 then find the value of m.
Solution:
p(y) = y³ – 5y² + 7y + m
Divisor = y + 2
Put y = – 2 in p(y)
∴ p(– 2) = (– 2) – 5(– 2)² + 7(– 2) + m
∴ p(– 2) = –8 – 20 – 14 + m
∴ p(– 2) = – 42 + m
By remainder theorem, p(– 2) = remainder
∴ Remainder = – 42 + m
But remainder = 50
∴ – 42 + m = 50
∴ m = 50 + 42
∴ m = 92
Ans: The value of m is 92.
(7) Use factor theorem to determine whether x + 3 is factor of x² + 2x – 3 or not.
Solution:
p(x) = x² + 2x – 3
Put x = – 3 in p(x)
∴ p(– 3) = (– 3)² + 2(– 3) – 3
∴ p(– 3) = 9 – 6 – 3
∴ p(– 3) = 0
As p(– 3) = 0,
By factor theorem, x + 3 is a factor of x² + 2x – 3.
(8) If (x – 2) is a factor of x³ – mx² + 10x – 20 then find the value of m.
Solution:
p(x) = x³ – mx² + 10x – 20
Put x = 2 in p(x)
∴ p(2) = (2)³ – m(2)² + 10(2) – 20
∴ p(2) = 8 – 4m + 20 – 20
∴ p(2) = 8 – 4m
As p(x – 2) is a factor of p(x),
∴ By factor theorem, p(2) = 0
∴ 8 – 4m = 0
∴ 8 = 4m
∴ m = \(\large \frac {8}{4}\)
∴ m = 2
Ans: The value of m is 2.
(9) By using factor theorem in the following examples, determine whether q(x) is a factor p(x) or not.
(i) p(x) = x³ – x² – x – 1, q(x) = x – 1
Solution:
p(x) = x³ – x² – x – 1
Put x = 1 in p(x)
∴ p(1) = (1)³ – (1)² – (1) – 1
∴ p(1) = 1 – 1 – 1 – 1
∴ p(1) = – 2
As p(1) ≠ 0,
By factor theorem, x – 1 is not a factor of x³ – x² – x – 1.
(ii) p(x) = 2x³ – x² – 45, q(x) = x – 3
Solution:
p(x) = 2x³ – x² – 45
Put x = 3 in p(x)
∴ p(3) = 2(3)³ – (3)² – 45
∴ p(3) = 54 – 9 – 45
∴ p(3) = 0
As p(3) = 0,
∴ By factor theorem, x – 3 is a factor of 2x³ – x² – 45.
(10) If (x³¹ + 31) is divided by (x + 1) then find the remainder.
Solution:
p(x) = x³¹ + 31
Divisor = x + 1
Put x = – 1 in p(x)
∴ p(–1) = (– 1)³¹ + 31
∴ p(–1) = – 1 + 31
∴ p(–1) = 30
By remainder theorem, p(– 1) = remainder
∴ Remainder = 30
(11) Show that m – 1 is a factor of m²¹ – 1 and m²² – 1.
Solution:
p(m) = m²¹ – 1
Put m = 1 in p(m)
∴ p(1) = (1)²¹ – 1
∴ p(1) = 1 – 1
∴ p(1) = 0
As p(1) = 0, by factor theorem, (m – 1) is a factor of m²1 – 1.
Now, p(m) = m²² – 1
Put m = 1 in p(m)
∴ p(1) = (1)²² – 1
∴ p(1) = 1 – 1
∴ p(1) = 0
As p(1) = 0, by factor theorem, (m – 1) is a factor of m²² – 1.
(12*) If x – 2 and x – \(\large \frac {1}{2}\) both are the factors of the polynomial nx² – 5x + m, then show that m = n = 2
Solution:
p(x) = nx² – 5x + m
Put x = 2 in p(x)
∴ p(2) = n(2)² – 5(2) + m
∴ p(2) = 4n – 10 + m
As x – 2 is a factor of p(x),
∴ By factor theorem p(2) = 0
∴ 4n – 10 + m = 0
∴ m + 4n = 10 … (i)
Now,
p(x) = nx² – 5x + m
Put x = \(\large \frac {1}{2}\) in p(x)
∴ p\(\large (\frac {1}{2})\) = n\(\large (\frac {1}{2})^2\) – 5\(\large \frac {1}{2}\) + m
∴ p\(\large (\frac {1}{2})\) = n × \(\large \frac {1}{4}\) – \(\large \frac {5}{2}\) + m
∴ p\(\large (\frac {1}{2})\) = \(\large \frac {n}{4}\) – \(\large \frac {5}{2}\) + m
As x – \(\large \frac {1}{2}\) is a factor of p(x),
∴ By factor theorem p\(\large (\frac {1}{2})\) = 0
∴ \(\large \frac {n}{4}\) – \(\large \frac {5}{2}\) + m = 0
∴ \(\large \frac {n}{4}\) – \(\large \frac {5\,×\,2}{2\,×\,2}\) + \(\large \frac {m\,×\,4}{4}\) = 0
∴ \(\large \frac {n}{4}\) – \(\large \frac {10}{4}\) + \(\large \frac {4m}{4}\) = 0
∴ \(\large \frac {n\,–\,10\,+\,4m}{4}\) = 0
∴ 4m + n = 10 …(ii)
Multiplying equation (ii) by 4 we get,
16m + 4n = 40 …(iii)
Subtracting equation (i) from (ii)
16m + 4n = 40
– m ± 4n = ±10
15m = 30
∴ m = \(\large \frac {30}{15}\)
∴ m = 2
Substituting m = 2 in equation (ii), we get,
4m + n = 10
∴ 4(2) + n = 10
∴ 8 + n = 10
∴ n = 10 – 8
∴ n = 2
∴ m = n = 2
Hence proved.
(13)
(i) If p(x) = 2 + 5x then p(2) + p(– 2) – p(1).
Solution:
p(x) = 2 + 5x
∴ p(2) = 2 + 5(2)
∴ p(2) = 2 + 10
∴ p(2) = 12 …(i)
∴ p(– 2) = 2 + 5(– 2)
∴ p(– 2) = 2 – 10
∴ p(–2) = – 8 …(ii)
∴ p(1) = 2 + 5(1)
∴ p(1) = 2 + 5
∴ p(1) = 7 …(iii)
∴ p(2) + p(– 2) – p(1)
= 12 + (– 8) – 7 …[from (i), (ii), (iii)]
= – 3
(ii) If p(x) = 2x² – 5\(\sqrt{3}\) x + 5 then p(5\(\sqrt{3}\)).
Solution:
p(x) = 2x² – 5\(\sqrt{3}\) x + 5
∴ p(5\(\sqrt{3}\)) = 2(5)² – 5\(\sqrt{3}\)(5\(\sqrt{3}\)) + 5
∴ p(5\(\sqrt{3}\)) = 2 × 75 – 75 + 5
∴ p(5\(\sqrt{3}\)) = 150 – 75 + 5
∴ p(5\(\sqrt{3}\)) = 80
Practice set 3.6
(1) Find the factors of the polynomials given below.
(i) 2x² + x – 1
Solution:
2x² + x – 1
= 2x² + 2x – x – 1
= 2x (x + 1) – 1 (x + 1)
= (x + 1) (2x – 1)
(ii) 2m² + 5m – 3
Solution:
2m² + 5m – 3
= 2m² + 6m – m – 3
= 2m (m + 3) – 1 (m + 3)
= (m + 3) (2m – 1)
(iii) 12x² + 61x + 77
Solution:
12x² + 61x + 77
= 12x² + 28x + 33x + 77
= 4x (3x + 7) + 11 (3x + 7)
= (3x + 7) (4x + 11)
(iv) 3y² – 2y – 1
Solution:
3y² – 2y – 1
= 3y² – 3y + y – 1
= 3y (y – 1) + 1(y – 1)
= (y – 1) (3y + 1)
(v) \(\sqrt{3}\) x² + 4x + \(\sqrt{3}\)
Solution:
\(\sqrt{3}\) x² + 4x + \(\sqrt{3}\)
= \(\sqrt{3}\) x² + 3x + x + \(\sqrt{3}\)
= \(\sqrt{3}\)x (x + \(\sqrt{3}\)) + 1(x + \(\sqrt{3}\))
= (x + \(\sqrt{3}\)) (\(\sqrt{3}\)x + 1)
(vi) \(\large \frac {1}{2}\)x² – 3x + 4
Solution:
\(\large \frac {1}{2}\)x² – 3x + 4
= \(\large \frac {1}{2}\)x² – \(\large \frac {3\,×\,2}{2}\)x + \(\large \frac {4\,×\,2}{2}\)
= \(\large \frac {1}{2}\)x² – \(\large \frac {6}{2}\)x + \(\large \frac {8}{2}\)
= \(\large \frac {x²\,–\,6x\,+\,8}{2}\)
= \(\large \frac {1}{2}\)(x² – 4x – 2x + 8)
= \(\large \frac {1}{2}\)[x(x – 4) –2(x – 4)]
= \(\large \frac {1}{2}\)(x – 4) (x – 2)
(2) Factorize the following polynomials.
(i) (x² – x)² –8 (x² – x) + 12
Solution:
(x² – x)² – 8 (x² – x) + 12
Let x² – x = m
∴ (x² – x)² – 8 (x² – x) + 12
= m² – 8m + 12
= m² – 6m – 2m + 12
= m (m – 6) – 2 (m – 6)
= (m – 6) (m – 2)
Resubstituting the value of m, we get,
(x² – x)² – 8 (x² – x) + 12
= (x² – x – 6) (x² – x – 2)
= [x² – 3x + 2x – 6] [x²– 2x + x – 2]
= [x(x – 3) + 2 (x – 3)] [x(x – 2) + 1 (x – 2)]
= (x – 3) (x + 2) (x – 2) (x + 1)
(ii) (x – 5)² –(5x – 25) – 24
Solution:
(x – 5)² – (5x – 25) – 24
= (x – 5)² – 5(x – 5) – 24
Let x – 5 = m
∴ (x – 5)² – 5(x – 5) – 24
= m² – 5m – 24
= m² – 8m + 3m – 24
= m (m – 8) + 3 (m – 8)
= (m – 8) (m + 3)
Resubstituting the value of m, we get,
(x – 5)² – (5x – 25) – 24
= (x – 5 – 8) (x – 5 + 3)
= (x – 13) (x – 2)
(iii) (x² – 6x)² – 8 (x² – 6x +8) – 64
Solution:
(x² – 6x)² – 8 (x² – 6x + 8) – 64
Let x² – 6x = m
∴ (x² – 6x)² – 8 (x² – 6x + 8) – 64
= m² – 8(m + 8) – 64
= m² – 8m – 64 – 64
= m² – 8m – 128
= m² – 16m + 8m – 128
= m (m – 16) + 8 (m – 16)
= (m – 16) (m + 8)
Resubstituting the value of m, we get,
(x² – 6x)² – 8 (x² – 6x + 8) – 64
= (x²– 6x – 16) (x²– 6x + 8)
= [x²– 8x + 2x – 16] [x²– 4x – 2x + 8]
= [x(x – 8) + 2 (x – 8)] [x(x – 4) – 2 (x – 4)]
= (x – 8) (x + 2) (x – 4) (x – 2)
(iv) (x² –2x +3) (x² –2x +5) – 35
Solution:
(x² – 2x + 3) (x² – 2x + 5) – 35
Let x² – 2x = m
∴ (x² – 2x + 3) (x² – 2x + 5) – 35
= (m + 3)(m + 5) – 35
= m² + 5m + 3m + 15 – 35
= m² + 8m – 20
= m² + 10m – 2m – 20
= m (m + 10) – 2 (m + 10)
= (m + 10) (m – 2)
Resubstituting the value of m, we get,
(x² – 2x + 3) (x² – 2x + 5) – 35
= (x² – 2x + 10) (x² – 2x – 2)
(v) (y + 2) (y – 3)(y + 8) (y + 3) + 56
Solution:
(y + 2) (y – 3) (y + 8) (y + 3) + 56
= (y – 3) (y + 8) (y + 2) (y + 3) + 56
= (y² + 5y – 24) (y² + 5y + 6) + 56
Let y² + 5y = m
∴ (m – 24) (m + 6) + 56
= m² + 6m – 24m – 144 + 56
= m² – 18m – 88
= m² – 22m + 4m – 88
= m (m – 22) + 4 (m – 22)
= (m – 22) (m + 4)
Resubstituting the value of m, we get,
(y + 2) (y – 3) (y + 8) (y + 3) + 56
= (y² + 5y – 22) (y² + 5y + 4)
= (y² + 5y – 22) (y² + 4y + y + 4)
= (y² + 5y – 22) [y(y + 4) + 1 (y + 4)]
= (y² + 5y – 22) (y + 4) (y + 1)
(vi) (y² + 5y) (y² + 5y –2) – 24
Solution:
(y² + 5y) (y² + 5y – 2) – 24
Let y² + 5y = m
∴ (y² + 5y) (y² + 5y – 2) – 24
= m (m – 2) – 24
= m² – 2m – 24
= m² – 6m + 4m – 24
= m (m – 6) + 4 (m – 6)
= (m – 6) (m + 4)
Resubstituting the value of m, we get,
∴ (y² + 5y) (y² + 5y – 2) – 24
= (y² + 5y – 6) (y² + 5y + 4)
= (y² + 6y – y – 6) (y² + 4y + y + 4)
= [y(y + 6) – 1 (y + 6)] [y (y + 4) + 1 (y + 4)]
= (y + 6) (y – 1) (y + 4) (y + 1)
(vii) (x – 3)(x – 4)² (x – 5) – 6
Solution:
(x – 3) (x – 4)² (x – 5) – 6
= (x – 3) (x – 4) (x – 4) (x – 5) – 6
= (x – 5) (x – 3) (x – 4) (x – 4) – 6
= (x² – 8x + 15) (x² – 8x + 16) – 6
Let x² – 8x = m
∴ (m + 15) (m + 16) – 6
= m² + 16m + 15m + 240 – 6
= m² + 31m + 234
= m² + 18m + 13m + 234
= m (m + 18) + 13 (m + 18)
= (m + 18) (m + 13)
Resubstituting the value of m, we get,
(x – 3) (x – 4)² (x – 5) – 6
= (x² – 8x + 18) (x² – 8x + 13)
Problem Set 3
(1) Write the correct alternative answer for each of the following questions.
(i) Which of the following is a polynomial ?
(A) \(\large \frac {x}{y}\)
(B) \(x^{\(\sqrt{2}\)}\) – 3x
(C) x⁻² + 7
(D) \(\sqrt{2}\)x² + \(\large \frac {1}{2}\)
Ans: Option (D) : \(\sqrt{2}\)x² + \(\large \frac {1}{2}\)
(ii) What is the degree of the polynomial \(\sqrt{7}\)?
(A) \(\large \frac {1}{2}\)
(B) 5
(C) 2
(D) 0
Ans: Option (D) : 0
(iii) What is the degree of the 0 polynomial?
(A) 0
(B) 1
(C) undefined
(D) any real number
Ans: Option (C) : undefined
(iv) What is the degree of the polynomial 2x² + 5x³ + 7 ?
(A) 3
(B) 2
(C) 5
(D) 7
Ans: Option (A) : 3
(v) What is the coefficient form of x³ – 1 ?
(A) (1, – 1)
(B) (3, – 1)
(C) (1, 0, 0, – 1)
(D) (1, 3, – 1)
Ans: Option (C) : (1, 0, 0, – 1)
(vi) p(x) = x² – 7\(\sqrt{7}\)x + 3 then p(7\(\sqrt{7}\)) = ?
(A) 3
(B) 7\(\sqrt{7}\)
(C) 42\(\sqrt{7}\) + 3
(D) 49\(\sqrt{7}\)
Ans: Option (D) : 49\(\sqrt{7}\)
(vii) When x = – 1, what is the value of the polynomial 2x³ + 2x ?
(A) 4
(B) 2
(C) – 2
(D) – 4
Ans: Option (A) : 4
(viii) If x – 1 is a factor of the polynomial 3x² + mx then find the value of m.
(A) 2
(B) – 2
(C) – 3
(D) 3
Ans: Option (C) : – 3
(ix) Multiply (x² – 3) (2x – 7x³ + 4) and write the degree of the product.
(A) 5
(B) 3
(C) 2
(D) 0
Ans: Option (A) : 5
(x) Which of the following is a linear polynomial?
(A) x + 5
(B) x² + 5
(C) x³ + 5
(D) x⁴ + 5
Ans: Option (A) : x + 5
(2) Write the degree of the polynomial for each of the following.
(i) 5 + 3x⁴
Ans: 4
(ii) 7
Ans: 0
(iii) ax⁷ + bx⁹ (a, b are constants.)
Ans: 9
(3) Write the following polynomials in standard form.
(i) 4x² + 7x⁴ – x³ – x + 9
Ans:
4x² + 7x⁴ – x³ – x + 9
Standard form : 7x⁴ – x³ + 4x² – x + 9
(ii) p + 2p³ + 10p² + 5p⁴ – 8
Ans:
p + 2p³ + 10p² + 5p⁴ – 8
Standard form : 5p⁴ + 2p³ + 10p² + p – 8
(4) Write the following polynomial in coefficient form.
(i) x⁴ + 16
Solution:
p(x) = x⁴ + 16
p(x) in index form = x⁴ + 0x³ + 0x² + 0x + 16
∴ p(x) in coefficient form = (1, 0, 0, 0, 16)
(ii) m⁵ + 2m² + 3m + 15
Solution:
p(x) = m⁵ + 2m² + 3m + 15
p(x) in index form = m⁵ + 0m⁴ + 0m³ + 2m² + 3m + 15
∴ p(x) in coefficient form = (1, 0, 0, 2, 3, 15)
(5) Write the index form of the polynomial using variable x from its coefficient form.
(i) (3, – 2, 0, 7, 18)
Solution:
The polynomial (3, – 2, 0, 7, 18) contains 5 coefficients
∴ The degree of the polynomial is 5 – 1 = 4
∴ The index form of the given polynomial is 3x⁴– 2x³ + 0x² + 7x + 18
(ii) (6, 1, 0, 7)
Solution:
The polynomial (6, 1, 0, 7) contains 4 coefficients
∴ The degree of the polynomial is 4 – 1 = 3
∴ The index form of the given polynomial is 6x³ + x² + 0x + 7
(iii) (4, 5, – 3, 0)
Solution:
The polynomial (4, 5, – 3, 0) contains 4 coefficients
∴ The degree of the polynomial is 4 – 1 = 3
∴ The index form of the given polynomial is 4x³ + 5x² – 3x + 0
(6) Add the following polynomials.
(i) 7x⁴ – 2x³ + x + 10 ; 3x⁴ + 15x³ + 9x² – 8x + 2
Solution:
7x⁴ – 2x³ + x + 10 + 3x⁴ + 15x³ + 9x² – 8x + 2
= 7x⁴+ 3x⁴– 2x³ + 15x³ + 9x² + x – 8x + 10 + 2
= 10x⁴+ 13x³ + 9x² – 7x + 12
(ii) 3p³q + 2p²q + 7 ; 2p²q + 4pq – 2p³q
Solution:
3p³q + 2p²q + 7 + 2p²q + 4pq – 2p³q
= 3p³q – 2p³q + 2p²q + 2p²q + 4pq + 7
= p³q + 4p²q + 4pq + 7
(7) Subtract the second polynomial from the first.
(i) 5x² – 2y + 9 ; 3x² + 5y – 7
Solution:
5x² – 2y + 9 – (3x² + 5y – 7)
= 5x² – 2y + 9 – 3x² – 5y + 7
= 5x² – 3x² – 2y – 5y + 9 + 7
= 2x² – 7y + 16
(ii) 2x² + 3x + 5 ; x² – 2x + 3
Solution:
2x² + 3x + 5 – (x² – 2x + 3)
= 2x² + 3x + 5 – x² + 2x – 3
= 2x² – x² + 3x + 2x + 5 – 3
= x² + 5x + 2
(8) Multiply the following polynomials.
(i) (m³ – 2m + 3)(m⁴ – 2m² + 3m + 2)
Solution:
(m³ – 2m + 3) (m⁴ – 2m² + 3m + 2)
= m³ (m⁴ – 2m²+ 3m + 2) – 2m (m⁴ – 2m² + 3m + 2) + 3 (m⁴ – 2m² + 3m + 2)
= m⁷ – 2m⁵ + 3m⁴ + 2m³– 2m⁵ + 4m³ – 6m² – 4m + 3m⁴ – 6m² + 9m + 6
= m⁷ – 4m⁵ + 6m⁴ + 6m³ – 12m² + 5m + 6
(ii) (5m³ – 2)(m² – m + 3)
Solution:
(5m³ – 2) (m² – m + 3)
= 5m³ (m² – m + 3) – 2 (m² – m + 3)
= 5m⁵ – 5m⁴ + 15m³– 2m² + 2m – 6
(9) Divide polynomial 3x³ – 8x² + x + 7 by x – 3 using synthetic method and write the quotient and remainder.
Solution:
Dividend = 3x³ – 8x² + x + 7
Dividend in coefficient form = (3, –8, 1, 7)
Comparing divisor x – 3 with x – a, we get, a = 3
Quotient in coefficient form is (3, 1, 4)
∴ Quotient = 3x² + x + 4, Remainder = 19
Ans: The quotient is 3x² + x + 4 and the remainder is 19.
(10) For which the value of m, x + 3 is the factor of the polynomial x³ – 2mx + 21 ?
Solution:
p(x) = x³ – 2mx + 21
Put x = – 3 in p(x)
∴ p(– 3) = (– 3)³ – 2m(– 3) + 21
∴ p(– 3) = – 27 + 6m + 21
∴ p(– 3) = 6m – 6
As (x + 3) is a factor of p(x),
∴ By factor theorem, p(–3) = 0
∴ 6m – 6 = 0
∴ 6m = 6
∴ m = \(\large \frac {6}{6}\)
∴ m = 1
Ans: For m = 1, x + 3 is the factor of the polynomial x³ – 2mx + 21.
(11) At the end of the year 2016, the population of villages Kovad, Varud, Chikhali is 5x² – 3y², 7y² + 2xy and 9x² + 4xy respectively. At the beginning of the year 2017, x² + xy – y², 5xy and 3x² + xy persons from each of the three villages respectively went to another village for education, then what is the remaining total population of these three villages?
Solution:
Total population of three villages at the end of year 2016.
= 5x² – 3y² + 7y² + 2xy + 9x² + 4xy
= 5x² + 9x²– 3y² + 7y² + 2xy + 4xy
= 14x² + 4y² + 6xy
Total population of three villages left for Education at the beginning of year 2017.
= x² + xy – y² + 5xy + 3x² + xy
= x² + 3x² – y² + xy + 5xy + xy
= 4x² – y² + 7xy
Remaining total population of three villages at the end of year 2017.
= 14x² + 4y² + 6xy – (4x² – y² + 7xy)
= 14x² + 4y² + 6xy – 4x² + y² – 7xy
= 14x² – 4x² + 4y² + y² + 6xy – 7xy
= 10x² + 5y² – xy
Ans: The remaining total population of these three villages is 10x² + 5y² – xy.
(12) Polynomials bx² + x + 5 and bx³ – 2x + 5 are divided by polynomial x – 3 and the remainders are m and n respectively. If m – n = 0 then find the value of b.
Solution:
p(x) = bx² + x + 5
Divisor = (x – 3)
Put x = 3 in p(x)
∴ p(3) = b(3)² + (3) + 5
∴ p(3) = 9b + 8
By Remainder theorem, p(3) = remainder
∴ Remainder = 9b + 8
∴ m = 9b + 8 … (I)
q(x) = bx³ – 2x + 5
Divisor = (x – 3)
Put x = 3 in q(x)
∴ q(3) = b(3)³ – 2(3) + 5
∴ q(3) = 27b – 6 + 5
∴ q(3) = 27b – 1
By Remainder theorem, q(3) = remainder
∴ Remainder = 27b – 1
∴ n = 27b – 1 … (ii)
Now, m – n = 0 …[Given]
∴ 9b + 8 – (27b – 1) = 0 [from (i) and (ii)]
∴ 9b + 8 – 27b + 1 = 0
∴ – 18b + 9 = 0
∴ – 18b = – 9
∴ 18b = 9
∴ b = \(\large \frac {9}{18}\)
∴ b = \(\large \frac {1}{2}\)
Ans: The value of b is \(\large \frac {1}{2}\)
(13) Simplify.
(8m² + 3m – 6) – (9m – 7) + (3m² – 2m + 4)
Solution:
(8m² + 3m – 6) – (9m – 7) + (3m² – 2m + 4)
= 8m² + 3m – 6 – 9m + 7 + 3m² – 2m + 4
= 8m² + 3m² + 3m – 9m – 2m – 6 + 7 + 4
= 11m² – 8m + 5
(14) Which polynomial is to be subtracted from x² + 13x + 7 to get the polynomial 3x² + 5x – 4?
Solution:
Let p(x) be subtracted from x² + 13x+ 7 to get 3x² + 5x – 4
∴ x² + 13x + 7 – p(x) = 3x² + 5x – 4
∴ x² + 13x + 7 – (3x² + 5x – 4) = p(x)
∴ x² + 13x + 7 – 3x² – 5x + 4 = p(x)
∴ – 2x² + 8x + 11 = p(x)
∴ p(x) = – 2x² + 8x + 11
Ans: – 2x² + 8x + 11 should be subtracted from x² + 13x + 7 to get 3x² + 5x – 4.
(15) Which polynomial is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10?
Solution:
Let A be added to 4m + 2n+ 3 to get 6m + 3n+ 10
∴ (4m + 2n + 3) + A = 6m + 3n + 10
∴ A = 6m + 3n + 10 – (4m + 2n + 3)
∴ A = 6m + 3n + 10 – 4m – 2n – 3
∴ A = 2m + n + 7
Ans: 2m + n + 7 should be added to 4m + 2n + 3 to get 6m + 3n + 10.