chapter 9.2

Exercise 9.2.1

1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

4x2-3x+7

y2+√2

3√t+t√2

y+2/y

x10+y3+t50

Polynomial

Not a Polynomial

2. Write the coefficients of x2 in each of the following:

coefficient of x2 should be assigned by every every term.

(i)2+x2+x : coefficient is .

(ii)2−x2+x3 : coefficient is .

(iii)π2x2+x : coefficient is π/.

(iv)√2x-1 : coefficient is .

3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Binomial -

Monomial -

4. Write the degree of each of the following polynomials.

(i) 5x3 + 4x2 + 7x : .

(ii) 4 – y2 : .

(iii) 5t – √7 : .

(iv) 3 : .

5. Classify the following as linear, quadratic and cubic polynomials.

(i) x2+x- .

(ii)x−x3- .

(iii)y+y2+4 - .

(iv)1+x - .

(v)3t - .

(vi)r2- .

(vii)7x3- .

Exercise 9.2.2

1. Find the value of the polynomial 5x – 4x2 + 3 at

(i)x = 0

Substitute 0 in the equation.

= 5()-4()^2+3

= 5(0)-4(0)^2+3 = -+3

= 0-0+3 = .

(ii) x = –1

Substitute -1 in the equation.

= 5()-4()^2+3

= -5-4(1)+3 = -5-+3.

= -5-4+3 = .

(iii) x = 2

Substitute 2 in the equation.

= 5()-4()^2+3

= 10-4(4)+3 = 10-+3

= 10-16+3 = .

2. Find p(0), p(1) and p(2) for each of the following polynomials.

(i)p(y)=y^2-y+1

Substitute 0,1,2 in equation 1.

p(0) = ()^2 -+1 = .

p(1) = ()^2 -+1 = .

p(2) = ()^2 -+1 = .

(ii)p(t)=2+t+2t^2-t^3

Substitute 0,1,2 in equation 2.

p(0)=2++2()^2-()^3 = 2++- = .

p(1)=2++2()^2-()^3 = 2++-=.

p(2)=2++2()^2-()^3 = 2++-=.

(iii)p(x)=x^3

Substitute 0,1,2 in equation 3.

p(0)=()^3 = .

p(1)=()^3 = .

p(2)=()^3 = .

(iv)p(x)=(x-1)(x+1)

Substitute 0,1,2 in equation 4.

p(x)=x2-1 using identity property (a+b)(a-b)=a2−b2

p(0) = ()^2-1 =

p(1) = ()^2-1 =

p(2) = ()^2-1 =

3.Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1, x = -(1/3)

: p(-1/3) = 3x()+1 = .

Therefore,-1/3 is a zero of p(x).

(ii) p(x) = 5x - π , x = 4/5

: p(4/5) = 5() - π

: π =22/7

: p(4/5) = 4-22/7 = .

Therefore,4/5 is a not zero of p(x).

(iii) p(x) = (x + 1)(x - 2), x = -1, 2

: p(-1) = (+1)(-2) = x = .

: p(2) = (+1)(-2) = x = .

Therefore,1 and -1 is are zeroes of p(x).

(iv) p(x) = lx + m, x = -(m/l)

p(-m/l) = lx()+m = + = 0.

Therefore,-(m/l) is a zero of p(x).

(v) p(x) = 3x2 - 1, x = -(1/√3), 2/√3

p(-1/√3) = 3 x ()^2-1

= 3 x () - 1 = - = .

p(2/√3) = 3x(2/√3)^2-1 = 3(4/3)-1 = 4-1 = .

Therefore,2/√3 is not a zero of p(x).

4. Find the zero of the polynomial in each of the following cases.

(i) p(x) = x + 5 = 0 : x = is the zero.

(ii) p(x) = x – 5 = 0 : x = is the zero.

(iii) p(x) = 2x+5 = 0 : x = is the zero.

(iv) p(x) = 3x – 2 = 0 : x = is the zero.

(v) p(x) = 3x = 0 : x = is the zero.

(vi) p(x) = cx + d, c ≠ 0, c, d are real numbers, cx+d = 0 : x = is the zero.

Exercise 9.2.3

1. Determine which of the following polynomials has (x + 1) a factor.

(i) x3+x2+x+1

x + 1 = 0 : x = -1.

Substitute -1 in equation 1.

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

= -1+1-1+1 = .

(ii)x4+x3+x2+x+1

x + 1 = 0 : x = -1.

Substitute -1 in equation 2.

= (-1)^4+(-1)^3+(-1)^2+(-1)+1

= 1-1+1-1+1 = .

(iii)x4+3x3+3x2+x+1

x + 1 = 0 : x = -1.

Substitute -1 in equation 3.

= (-1)^4+3(-1)^3+3(-1)^2+(-1)+1

= 1+3(-1)+3(1)-1+1

= 1-3+3-1+1 =

x + 1 = 0 : x = -1.

Substitute -1 in equation 4.

= (-1)^3-(-1)^2-(2+√2)(-1)+√2

= -1-1-(-2-√2)+√2

= -1-1+2+√2+√2

= -2+2+2√2 = √2.

2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases.

(i)2x3+x2−2x−1,get(x)=x+1

(i)2x3+x2−2x−1

p(-1)=0

Substitute -1 in equation 1.

p(-1)=2(-1)^3+(-1)^2-2(-1)-1.

p(-1)=2(-1)+1+2-1

p(-1)=-2+1+2-1 = .

(ii)x3+3x2+3x+1,get(x)=x+2

(ii)x3+3x2+3x+1

p(-2)=0

Substitute -2 in equation 2.

p(-2)=(-2)^3+3(-2)^2+3(-2)+1.

p(-2)=-8+3(4)+3(-2)+1

p(-2)=-8+12-6+1 = .

3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases.

(i)p(x)=x2+x+k

p(x) = +k = 1+1+k = +k

p(1) = 0, 2 + k = 0, k = .

(ii)p(x) =

p(1)=212+k(1)+√2

p(1) = +k+√2

p(1)=0, 2+k+√2=0, k=-√2.

4.Factorise:

(i)12x2−7x+1

= 4x3x−1−13x−1

=(3x-1)(4x-1)

(ii)2x2+7x+3

= 2x2+6x+x+3

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

(iii)6x2+5x-6

= 6x2+9x-4x-6

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

(iv)3x2-x-4

= 3x2-4x+3x-4

= (3x-4)(x+1)

5.Factorise.

(i)x3-2x2-x+2

= x2(x-2)-1(x-2)(Common terms a side)

=(x-2)x2−1(Take x-2 as common)

(ii)x3-3x2-9x-5

= Here −3x2=−5x2+2x^2.

= Simplify x2-−5x2+2x2-10x+x-5.

= Taking x2,2x and 1 as common x2(x-5)+2x(x-5)+1(x-5) ⇒ (x-5)(x2+2x+1).

Apply a2+2ab+b2=a+b2 ⇒ As, a = x and b = 1.

Thus, ⇒

(iii)x3+13x2+32x+20

p(x)=x3+13x2+32x+20

Check p(x)=0,1,-1

x	x3+13x2+32x+20	Yes (or) No
0	03+1302+32(0)+20=20
1	13+1312+32(1)+20=60
-1	−13+13−12+32(-1)+20=0

Hence x+1 is a factor of p(x).

= x3+13x2+32x+20 ⇒ (x+1)x2+12x+20.

We factor the quadratic x2+12x+20, we look for two numbers that multiply to 20 and add to 12.

so,we can write : x2+12x+20 ⇒ (x+10) (x+2)

The complete factorization of the given polynomial is : x3+13x2+32x+20 ⇒ .

(iv)2y3+y2-2y-1

Common terms : y2(2y-1)-1(2y+1) ⇒ (2y+1)y2−1

: (2y+1)y2−1 ⇒ (2y+1)y2−12.

: (2y+1)((y+1)(y-1)) Apply a2-b2=(a+b)(a-b).

Exercise 9.2.4

1. Use suitable identities to find the following products.

Match the following

(x+4)(x+10)

(x+8)(x-10)

(3x+4)(3x-5)

(y^2) + (3/2)(y^2 -3/2)

(3-2x)(3+2x)

y^4-9/4

x^2-2x-80

9-4x^2

x^2+14x+40

9x^2-3x-20

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Glossary

chapter 9.2

Exercise 9.2.1

Exercise 9.2.2

Exercise 9.2.3

Exercise 9.2.4