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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)​

Respuesta :

Answer:

C. [tex](x-3)(x+6)[/tex]

Step-by-step explanation:

I think the equation meant [tex]x^2+3x-18[/tex]?

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

[tex]ax^2+bx+c=(x+w)(x+v)[/tex]

ONLY if:

[tex]w+v=b\\wv=c[/tex]

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.

[tex]1+(-18)\neq 3\\2+(-9)\neq 3\\3+(-6)\neq 3\\6+(-3)=3\\9+(-2)\neq 3\\18+(-1)\neq 3\\[/tex]

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

Replace them into the factored form:

[tex](x+w)(x+v)\\(x+6)(x-3)\\(x-3)(x+6)[/tex]

ROTRUY

Answer:

x²+3x-18 = (x-3)(x+6)

Step-by-step 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 [tex]x_{1} =\frac{-b+\sqrt{b^{2}-4ac } }{2a}[/tex] and [tex]x_{2} =\frac{-b-\sqrt{b^{2}-4ac } }{2a}[/tex]

(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:

[tex]x_{1} =\frac{-3+\sqrt{9+72}}{2}=\frac{-3+\sqrt{81}}{2}=\frac{-3+9}{2}=\frac{6}{2}=3[/tex]

[tex]x_{2} =\frac{-3-\sqrt{9+72}}{2}=\frac{-3-\sqrt{81}}{2}=\frac{-3-9}{2}=\frac{-12}{2}=-6[/tex]

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)

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