Question

find the zeros of the quadratic polynomial X^2 + 7x + 10​

Answer 1

Answer:

$x {}^{2} + 7x + 10 \\ = x {}^{2} + 5x + 2x + 10 \\ = x(x + 5) + 2(x + 5) \\ = ( x + 5)(x + 2)$

Answer 2

Given,

p(x) = x^2+7x+10

By factorization method, we obtain.......

x^2+7x+10

x^2+5x+2x+10

x(x+5)+2(x+5)

(x+5)(x+2).............is the factors of polynomial x^2+7x+10

Let.....

x+5 = 0

x = -5

Putting the value of x = -5, we get.....

p(x) = x^2+7x+10

p(-5) = (-5)^2+7(-5)+10

= 25-35+10

= -10+10

= 0

Let......

x+2 = 0

x = -2

Putting the value of x = -2, we get......

p(x) = x^2+7x+10

p(-2) = (-2)^2+7(-2)+10

= 4-14+10

= -10+10

= 0

Hence, -5 and -2 are the zeros of p(x).

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