Chapter 2: Polynomials

4x² – 3x – 1

Splitting the middle term, we get,

4x²-4x+1x-1

Taking the common factors out, we get,

4x(x-1) +1(x-1)

On grouping, we get,

(4x+1)(x-1)

So, the zeroes are,

4x+1= 0⇒ 4x=-1 ⇒x= (-1/4)

(x-1) = 0 ⇒ x=1

Therefore, zeroes are (-1/4) and 1

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x²

α + β = – b/a

1 – 1/4 = – (- 3)/4 = ¾

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

1(- 1/4) = – ¼

– 1/4 = – 1/4

3x² + 4x – 4

Splitting the middle term, we get,

3x² + 6x – 2x – 4

Taking the common factors out, we get,

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

On grouping, we get,

(x+2)(3x-2)

So, the zeroes are,

x+2=0 ⇒ x= -2

3x-2=0⇒ 3x=2⇒x=2/3

Therefore, zeroes are (2/3) and -2

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x²

α + β = – b/a

– 2 + (2/3) = – (4)/3

= – 4/3 = – 4/3

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

Product of the zeroes = (- 2) (2/3) = – 4/3

5t² + 12t + 7

Splitting the middle term, we get,

5t² +5t + 7t + 7

Taking the common factors out, we get,

5t (t+1) +7(t+1)

On grouping, we get,

(t+1)(5t+7)

So, the zeroes are,

t+1=0 ⇒ y= -1

5t+7=0 ⇒ 5t=-7⇒t=-7/5

Therefore, zeroes are (-7/5) and -1

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x²

α + β = – b/a

(- 1) + (- 7/5) = – (12)/5

= – 12/5 = – 12/5

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(- 1)(- 7/5) =  7/5

7/5 =  7/5

t³ – 2t² – 15t

Taking t common, we get,

t ( t² -2t -15)

Splitting the middle term of the equation t² -2t -15, we get,

t( t² -5t + 3t -15)

Taking the common factors out, we get,

t (t (t-5) +3(t-5)

On grouping, we get,

t (t+3)(t-5)

So, the zeroes are,

t=0

t+3=0 ⇒ t= -3

t -5=0 ⇒ t=5

Therefore, zeroes are 0, 5 and -3

Verification:

Sum of the zeroes = – (coefficient of x²) ÷ coefficient of x³

α + β + γ = – b/a

(0) + (- 3) + (5) = – (- 2)/1

= 2 = 2

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x³

αβ + βγ + αγ = c/a

(0)(- 3) + (- 3) (5) + (0) (5) = – 15/1

= – 15 = – 15

Product of all the zeroes = – (constant term) ÷ coefficient of x³

αβγ = – d/a

(0)(- 3)(5) = 0

1. = 0

2x² +(7/2)x +3/4

The equation can also be written as,

8x²+14x+3

Splitting the middle term, we get,

8x²+12x+2x+3

Taking the common factors out, we get,

4x (2x+3) +1(2x+3)

On grouping, we get,

(4x+1)(2x+3)

So, the zeroes are,

4x+1=0 ⇒ x = -1/4

2x+3=0 ⇒ x = -3/2

Therefore, zeroes are -1/4 and -3/2

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x²

α + β = – b/a

(- 3/2) + (- 1/4) = – (7)/4

= – 7/4 = – 7/4

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(- 3/2)(- 1/4) = (3/4)/2

3/8 = 3/8