Question

P(x)=x^2+7x+12
What are the numbers that can be 'x' if p (x) = 0?

Answer 1

Step-by-step explanation:

0=x²+7x+12

0=x²+(4+3)x+12

0=x²+4x+3x+12

0=x(x+4)+3(x+4)

0=(x+4)(x+3)

numbers can be x are -4 and -3

x+4=0

x=-4

x+3=0

x=-3

Answer 2

Answer:

The values of x are -2, -6

Step-by-step explanation:

Here we have been mentioned to find the value of the following above-given quadratic equation by solving it by using the factorization method.

We have to find the value of x if P(x) = 0

Therefore we have the following:

$P(x) = x^{2} + 7x + 12$

⇒ $x^{2} + (6+2) x + 12 = 0$

⇒ $x^{2} + 6x +2x+12 = 0$

⇒ $x(x+6) + 2(x+6) = 0$

⇒ $(x+2)(x+6)=0$

⇒ x = -2, -6

The values of x are -2 , -6