Question

Please answer, I haven’t no clue how to work it out

Answer 1

Answer:

c.) $(x-2)$

Step-by-step explanation:

The given expression is written in the form $ac^2+bx+c$. To factor, find two numbers whose sum is b and whose product is c:

_×_=-18

_+_=7

Use -2 and 7:

-2×9=-18

-2+9=7

Substitute the two numbers for b:

$x^2-2x+9x-18$

$(x^2-2x)+(9x-18)$

Factor out common terms. The common term in the first parentheses is x and the common term in the second is 9:

$x(\frac{x^2-2x}{x} )+9(\frac{9x-18}{9})\\\\x(x-2)+9(x-2)$

Cancel out one of the parentheses since they have the same terms and plug the factored numbers in:

$(x-2)(x+9)$

:Done

Check Your Work:

To see if the given factored form is true, simplify it using FOIL:

First, Outside, Inside, Last

$(x-2)(x+9)$

Multiply the first terms:

$x*x=x^2$

Multiply the terms on the outside:

$x*9=9x\\\\x^2+9x$

Multiply the inside terms:

$-2*x=-2x\\\\x^2+9x-2x$

Multiply the last terms:

$-2*9=-18\\\\x^2+9x-2x-18$

Combine like terms:

$x^2+(9x-2x)-18\\\\x^2+7x-18$

Therefore, the factored form is true.

Answer 2

Answer:

C. x-2

Step-by-step explanation:

x² + 7x - 18 = 0

the factor must be p and q from the form.of

(x+p)(x+q) = 0

then find the value of p and q by this rules

p + q = 7

p x q = -18

p = 9

q = -2

(x + 9)(x + (-2)) = 0

(x+9)(x-2) = 0