Factoring

[swimaria96]

I am in geometry righy now, but my teacher gave a algebra review sheet for homework. I completly forget how to do this. An example for a problem is 3x^2-20x-63. Could someone please explain how to factor this completly?

 

[Loren]

There are several different techniques for factoring a second degree trinomial. One of the most commonly used is FOIL in reverse. It goes something like this.

Assume that (ax+b)(cx+d) = 3x[sup:1j1ih1ra]2[/sup:1j1ih1ra]-20x-63

You have to find a, b, c and d such that....

1) ac = 3.
2) ad + bc = -20
and
3) bd = -63.

From 1) we know that either a=1 and c=3 or a=3 and c=1.
From 3) we have these possibilities.
b=1 and d=-63 or b=-63 and d=1
b=3 and d=-21 or b=-21 and d=3
b=7 and d=-9 or b=-9 and d=7.
I can't think of any other possibilities.

Now, you have to make the choices so that ad + bc = -20.

 

[soroban]

Hello, swimaria96!

Factor: .\(\displaystyle 3x^2-20x-63\)

There are several methods available . . . I'll show you my favorite.

The middle term has the coefficient \(\displaystyle \text{-}20.\)
We want a sum which is the middle coefficient, \(\displaystyle \text{-}20\)

Multiply the first coefficient by the last coefficient: .\(\displaystyle (3)(-63) \:=\:\text{-}189\)

\(\displaystyle \text{Factor -}189\text{ into all possible pairs: }\;\begin{Bmatrix}(\pm1)(\mp189) \\ (\pm3)(\mp63) \\ (\pm7)(\mp27) \\ (\pm9)(\mp21) \end{Bmatrix}\)

\(\displaystyle \text{The pair with a sum of }\text{-}20\text{ is: }\+7)(\text{-}27)\)


\(\displaystyle \text{Use them as coefficients for the middle term: }\;3x^2 \;\overbrace{+\;7x - 27x}^{-20x} -\; 63\)

. . . . . . . . . . . . . . . . .\(\displaystyle \text{Factor by grouping: }\;x(3x + 7) - 9(3x + 7)\)

. . . . . . . . . \(\displaystyle \text{Factor out the common factor: }\quad\;\;(3x+7)(x-9)\)