Question

Factor this polynomial completely x2 + 3x – 18 A. (x-2)(x + 9) B. (x+3)(x-6) оо C. (x - 3)(x+6) D. (x - 3)(x-6)​

Answer 1

x²+3x-18 = (x-3)(x+6)

Explanation:

Polynomial: x²+3x-18

You can see that this is in the form of a quadratic equation, writing it in equation form we get: x²+3x-18 = 0

As this is a standard quadratic equation, we just need to find the two solutions and fill them in into the formula ax²+bx+c = (x-x₁)(x-x₂)

where
`x_(1) =\frac{-b+\sqrt{b^(2)-4ac } }{2a}` and
`x_(2) =\frac{-b-\sqrt{b^(2)-4ac } }{2a}`

(or the other way around, doesn't matter which one is x₁ and x₂)

Filling in the numbers (a=1, b=3, c=-18) we find that the equation has the following solutions:

`x_(1) =(-3+√(9+72))/(2)=(-3+√(81))/(2)=(-3+9)/(2)=(6)/(2)=3`

`x_(2) =(-3-√(9+72))/(2)=(-3-√(81))/(2)=(-3-9)/(2)=(-12)/(2)=-6`

This means that we can just fill them in into the formula ax²+bx+c = (x-x₁)(x-x₂):

We get that: x²+3x-18 = (x-3)(x+6)

Answer 2

C.
`(x-3)(x+6)`

Explanation:

I think the equation meant
`x^2+3x-18`?

Anyways, to factor these kinds of quadratic, keep into consideration:

`ax^2+bx+c=(x+w)(x+v)`

ONLY if:

`w+v=b\\wv=c`

Start off by finding factors of c, which in this case, -18:

±(1, 2, 3, 6, 9, 18)

If one of the numbers is negative then the other number must be positive.

Find which two factors will sum up to b, which in this case, is 3.

`1+(-18)\\eq 3\\2+(-9)\\eq 3\\3+(-6)\\eq 3\\6+(-3)=3\\9+(-2)\\eq 3\\18+(-1)\\eq 3\\`

The only two factors that work are 6 and -3.

Replace them into the factored form:

`(x+w)(x+v)\\(x+6)(x-3)\\(x-3)(x+6)`