CBSE Solutions For Class 10 Mathematics Chapter 2 Polynomials

CBSE Solutions For Class 10 Mathematics Chapter 2

Question 1. x²+9x+20

Solution:

Let P(x) = = x²+9x+20

= x²+5x+4x+20

= x(x+5)+4(x+5)

(x+4)(x+5)

P(x)=0

(x+4)(x+5)=0

x+4=0 or x+5=0

x=-4 x=-5

Zeros of P(x) are -4 and -5 \(=\frac{-9}{1}=\frac{- \text { Coefficient of } x}{\text { coefficient of } x^2}\)

Now, Sum of Zeros = -4+(-5) \(\frac{20}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

Question 2. x² – 9

Solution:

Let P(x) = x² = 9 = x²=-32 = (x-3)(x+3)

P(x)=0

(x-3)(x+3)=0

x-3=0 Or x+3=0

x=3 x=-3

Zeros of p(x) are 3 and -3

Now, Sum of Zeros = 3+(-3)=0 = \(\frac{-0}{1}=-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^2}\)

and Product of Zeros = (3)(-3) = −9 = \(\frac{-9}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

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Question 3. Find the quadratic polynomials, the Sum of whose Zeroes is 17 and the product is 60. Hence, find the Zeroes of the polynomial.

Solution:

Let and B be the Zeroes of the polynomial P(X).

Given that X+B=17 and αβ=60

Now, P(x) = x²= (α+β)γ + αβ

= x²= 17x+60

= x²=12x-5x+60

= x(x-12)-5(x-12)

= (x-5)(x-12)

There may be So many different polynomials that satisfy the given Condition. The general equation quadratic polynomial will be k(x2=-17x+60), where k = 0

P(x)=0

(x-5) (x-12) = 0

(x-5)=0 or (x-2)=0

2=5 Or x=12

Zeros are 12 and 5.

Question 4. Find a quadratic polynomial, the Sum of whose Zeros is 7 and the product is -60. Hence, verify the relation between Zeros and Coefficients of the polynomial.

Solution:

Let and B be the Zeros of the polynomial P(x).

Given that α+β=ϒ and αβ = -60

Now, P(x) = x² – (α+β) γ+ αβ

= x²-7x-60

= x²-12x+5x-60

= x(x-2)+5(x-12)

= (x-12)+(x-5)

There may be so many different polynomials which satisfy the given Condition. The general quadratic polynomial will be k (x²-7x-60), where k = 0.

P(x)=0

(x-12) (x+5)=0

(x-12)=0

x=12 or (x+5)=0

x=-5

Zeros are 12 and -5.

Question 5. If the product of Zeroes of the polynomial 30+ 5x+k is 6, find the value of k.

Solution:

Given polynomial = 31751+k

Product of Zeroes = \(\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

6= \(\frac{k}{3}\)

= 6×3=k

⇒ k = 18

Question 6. If the Sum of Zeroes of the polynomial x²+2x-12 is 1, find the value of k.

Solution:

Given polynomial = x²+2kx-12

Sum of Zeroes = \(-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^2}\)

1 = \(\frac{-2 k}{1}\)

1 = -2K

⇒ \(k=\frac{-1}{2}\)

Question 7. If x = \(\frac{5}{3}\) and x = \(\frac{-1}{2}\) are the Zeroes of the polynomial ax=7x+b, then find the values of a and b.

Solution: Let P(x) = ax27x+b

⇒ \(x=\frac{5}{3} \text { and } x=\frac{-1}{2}\) are zeroes of p(x)

⇒ \(P\left(\frac{5}{3}\right)=0\)

⇒ \(a\left(\frac{5}{3}\right)^2-7\left(\frac{5}{3}\right)+b\)

⇒ \(\frac{25 a}{9}-\frac{35}{3}+b\)

⇒ \(b=\frac{-25 a}{9}+\frac{35}{3}\)

⇒ \(P\left(\frac{-1}{2}\right)=a\left(\frac{-1}{2}\right)^2-7\left(\frac{-1}{2}\right)+b\)

⇒ \(\frac{a}{4}+\frac{7}{2}+b\)

⇒ \(\frac{a}{4}+\frac{7}{2}-\frac{25 a}{9}+\frac{35}{3}=0\)

⇒ \(\frac{a}{4}-\frac{25 a}{9}=-\frac{7}{2}-\frac{35}{3}\)

⇒ \(\frac{9 a-100 a}{36}=\frac{-21-70}{6}\)

⇒ \(-\frac{91 a}{36}=\frac{-91}{6}\)

⇒ \(\frac{91 a}{6}=91\)

91a = 546

⇒ \(a=\frac{546}{91}\)

a = 6

⇒ \(b=\frac{-25(6)}{9}+\frac{35}{3}\)

⇒ \(b=\frac{-25(6)}{9}+\frac{35}{3}\)

⇒ \(b=\frac{-150+105}{9}\)

⇒ \(b=\frac{-45}{9}\)

b = -5

Question 8. Verify that 1,-2, 4 are Zeros of the Cubic polynomial x³-3x²-6x+8. Also, Verify the relation between Zeroes and Coefficients of the polynomial.

Solution:

Here, P(x) = x³-3x²-6x+8

P(1) = (1)² = 3(1)² – 6(1) +8 = 1-3-6+ 8 = −9+9=0

P(-2) = (2) ³ -3(-2)=6(-2)+8=-8-12+12+8 = 0

P(4) = (4)³-3(4) -6(4) +8 = 64-48-24+8 = 0

1-2 and 4 are Zeroes of P(x).

Now, α+B+= 1-2+4=3 = \(\frac{-3}{1}=-\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^3}\)

XB+B++√α = (1)(-2)+(-2) (4)+(4)(1) = −2-8+4

= -10+4

⇒\(\frac{-6}{1}=\frac{\text { Coefficient of } x}{\text { Coefficient of } x^3}\)

and LB7 = (1)(-2) (4) =\(-\frac{8}{1}=\frac{8}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^3}\)

Question 9. Verify that 2-4. and are zeroes of the Cubic polynomial 3x ³ +5x² = 26x+8. Also, verify the relation between Zeroes and Coefficients of the polynomial.

Solution:

Here, P(x)=3x³ +5 x 2 – 26x+8

P(2) = 3(2)3 +5(2)2 = 26(2)+8=24+20-52+8 = 52-52 = 0

P(-4)=3(-4)3 + 5(-4)=26(-4)+8=-(92+80+104+8=-192+192=0

⇒ \(P\left(\frac{1}{3}\right)=3\left(\frac{1}{3}\right)^3+5\left(\frac{1}{3}\right)^2-26\left(\frac{1}{3}\right)+8=\frac{3}{27}+\frac{5}{9}-\frac{26}{3}+8=\frac{3+15-234+216}{27}\)

⇒ \(=\frac{234-234}{27}=0\)

2,-4. and — are Zeroes of P(x),

Now \(\text { , } \alpha+\beta+\gamma=2-4+\frac{1}{3}=-2+\frac{1}{3}=\frac{-6+1}{3}=-\frac{5}{3}=\frac{5}{3}=-\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^3}\)

⇒ \(\alpha \beta+\beta \gamma+\gamma \alpha=(2)(-4)+(-4)\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)(2)=-8-\frac{4}{3}+\frac{2}{3}=\frac{-28+2}{3}=\frac{-26}{3}\)

⇒ \(=\frac{- \text { Coefficient of } x^2}{\text { coefficient of } x^3}\)

and \(\alpha_\beta \beta=(2)(-4)\left(\frac{1}{3}\right)=-\frac{-8}{3}=\frac{8}{3}=\frac{\text { Constant term }}{\text { Coefficient of } x^3}\)

Question 10. Find a Cubic polynomial whose Zeroes are 5, 6, and -4.

Solution:

Let α= 5, β=6 and γ=-4

α+β+ γ = 5+6-4 =) ||- 4 = 7

αB+ βγ + γα = 5(6)+6(-4)+(-4) (5)

= 30-24-20

= 30-44

= -14

XB = 5(6)(u)

= -120

Cubic Polynomial = x³-(x+β++) x² + (αβ+ βγ + γα) x-px

= x² – (7) x2 + (-14)x-(-120)

= x³-7x² = 14x+120

Question 11. Find a Cubic polynomial whose Zeroes are 11 and -1.

Solution:

let α= \(\frac{1}{2},\),B = 1 and 2 =-1

⇒ \(\alpha+\beta+\gamma=\frac{1}{2}+1-1=\frac{1}{2}\)

⇒ \(\alpha_\beta+\beta^1+\gamma \alpha=\frac{1}{2}(1)+(1)(-1)+(-1)\left(\frac{1}{2}\right) \Rightarrow \frac{1}{2}-1-\frac{1}{2} \Rightarrow-1\)

⇒ \(\alpha \beta \gamma=\frac{1}{2}(1)(-1) \Rightarrow \frac{-1}{2}\)

Cubic polynomial = x³= (α+β+γ) x2 + (αβ+ βγ + γα)x-αßγ

⇒ \(x^3-\frac{1}{2} x^2+(-1) x-\left(-\frac{1}{3}\right)\)

⇒ \(x^3-\frac{1}{2} x^2-x+\frac{1}{2}\)

⇒ \(2 x^3-x^2-2 x+1\)

Question 12. Find the quotient and remainder in each of the following and verify the division algorithm:

1. P(x) = x²=14x²+2x-1 is divided by g(x)=x+2

Solution:

Now, quotient = x2=6x+14, remainder = -29

dividend = x³-4x²+2x-1 and divisor = x+ 2

and quotient x divisor + remainder = (x²-6x+(4)(x+2)-29

= x²-6x²+14x+2x²=1271+28-29

= x²-4x² +2x-1

= dividend

2. P(x) = x² + 2x²= x + 1 is divided by g(x) = x²+1

Solution:

Now, quotient = x²+1, remainder = -x, dividend = x² + 2x = x +1 and

divisor = x+1

and quotient divisor + remainder = (x² + 1)(x² + 1) – α = ) x + x2 + x² + 1-x

x 4 + 2x = x + 1 =) dividend

Question 13. Actual division shows that x+2 is a factor of x³ + 4x²+3x-2.

Solution:

Question 14. If I am a zero of the polynomial x²-4x²=7x+10, find its other two Zeroes

Solution:

let P(x)=x³=_472-70+10

x= 1 is a zero of P(x)

(x-1) is a factor of P(x)

P(x) = x3- 4x²-7x+ 10 =) (x-1) (x=3x-10)

(x-1) (x²+2x-5x-10)

(x-1) ((x(x+2)-5(x+2))

(x-1)(x+2)(x-5)

Now, P(x)=0

⇒ (x-1)(x+2)(x-5)=0

⇒ α-1=0 or x+2=0 or x-5=0

x=1 or x=-2 or x=5

Hence, other Zeroes are -2 and 5

Question 15. If Land -2 are two Zeroes of the polynomial x⁴+x³=11x=9x+18, find the other two Zeroes.

Solution:

Let P(x) = x² + x² – 1172=9x+18

x=1, x=-2 is a zero of p(x)

(x-1) (x+2) is a factor of P(x)

x²+2x-x-2 =) x²+x-2 is a factor of P(x)

p(x0 = x4 = x3 – 11×2 – 9x + 18 = (x-1)(x+2)(x2-9)

= (x-1)(x+2)(x²-32)

= (x-1)(x+2)(x+3)(x-3)

Now P(x)

(x-1)(x+2)(x+3)(x-3)=0

= (x-1)(x+2)(x+3)(x-3)

⇒ x-1=0 x+2=0

⇒ x+3=0 X-3=0

⇒ x=1 x=-2 x=-3 X=3

Hence, other Zeroes are 3 and -3.

Question 16. Find all Zeroes of x + x3-23x=-3x+60, if it is given that two of its Zeroes are √3 and -√3.

Solution:

Let P(x) = x2+x3-28x=-3x+60

√3 and -√3 are Zeroes of P(x).

(x−√3)(x+√3) = x2= 3 is a factor of P(x).

P(x) = x²+x³-x²=3x+60 = (x²=3)(x²+11-20)

= (x²-3) [x²+5x-4x-20]

=(x²-3)(x(x+5)-4(x+5))

= (x²-3)(x-4)(x+5)

The other Zeroes are given by

x-4=0 or x+5=0

⇒ x=4 Οr x=-5

Hence, other Zeroes are 4 and -5.

Question 17. Find Zeroes of the polynomial f(x) = x² – 13x² + 32x – 60, if it is given that the Product of its two Zeroes is 10.

Solution: Let α, B, be Zeroes of the given polynomial p(x), Such that &p=10-)(1)

⇒ \(\alpha+\beta+\gamma=\frac{-(-13)}{1}=13 \longrightarrow(2)\)

⇒ \(\alpha \beta+\beta 1+\alpha \gamma=\frac{32}{1}=32 \longrightarrow(3)\)

⇒ \(\alpha \beta \gamma=-\frac{(-60)}{1}=60 \longrightarrow(4)\)

From (1) and (4)

10s=60

⇒ \(\gamma=\frac{60}{10} \Rightarrow \gamma=6 \rightarrow(5)\)

Put 7=6 in (2), we get α+3 +6 = 13

α+B=7

Now, (α-B)² = (x+3)=4xß

= (7)=4(10)

= 49-40

= 9

α-B = ± 3 —– (6)

Solving (5) and (6), we get

α=2,B=5 or α=2, B=5 and 1=6.

So, Zeroes are 2,5 and 6.

Question 18. What must be added to P(x) = 4x² – 5x = 39x = 46x = -2, so that the resulting Polynomial is divisible by g(x) = 4x² + 7x + 2 ?

Solution:

P(x) = 4×4 -5x³-39x²– 46x-2

9(x)= 4x²+7x+2