Question

Find the zeroes of the quadratic polynomial. (x-4)(x+7).​

Answer 1

Answer:

p(x)=x2+7x+10

comparing with p(x)=ax2+bx+c

a=1,b=7,c=10

p(x)=x2+7x+10

⇒x2+5x+2x+10

⇒x(x+5)+2(x+5)

⇒(x+2)(x+5)

Let p(x)=0

⇒(x+2)(x+5)=0

x=−2,5

α=−2,β=−5

a−b=α+β⇒1−(7)=−7

ac=α⋅β⇒110=10

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Answer 2

Answer:

4 and -7

Explanation:

(x-4)(x+7) = 0

So, either of the factors have to be 0.

x-4 = 0

=> x = 4

x+7 = 0

=> x = -7

So, the zeroes of the polynomial are 4 and -7.