randRangeNonZero(-9, 9) randRangeNonZero(-9, 9) randRangeNonZero(-5, 5) E + A * F E * F

Factor the expression below completely. All coefficients should be integers.

`A` `x^2+` `B` `x+` `C`

(Ax + E)(x + F)

This expression is in the form `\blue{A}x^2 + \green{B}x + \pink{C}`. You can factor it by grouping.

$("#question-a").addClass("hint_blue"); $("#question-b").addClass("hint_green"); $("#question-c").addClass("hint_pink");

First, find two values, `a` and `b`, so:

`\qquad \begin{eqnarray} \purple{ab} &=& \blue{A}\pink{C} \\ \purple{a} + \purple{b} &=& \green{B} \end{eqnarray}`

In this case:

`\qquad \begin{eqnarray} \purple{ab} &=& \blue{(A)}\pink{(C)} &=& A * C \\ \purple{a} + \purple{b} &=& & & \green{B} \end{eqnarray}`

In order to find `\purple{a}` and `\purple{b}`, list out the factors of `A * C` and add them together. Remember, since `A * C` is negative, one of the factors must be negative. The factors that add up to `\green{B}` will be your `\purple{a}` and `\purple{b}`.

When `\purple{a}` is `\purple{E}` and `\purple{b}` is `\purple{A * F}`:

`\qquad \begin{eqnarray} \purple{ab} &=& (\purple{E})(\purple{A * F}) &=& E * A * F \\ \purple{a} + \purple{b} &=& \purple{E} + \purple{A * F} &=& E + A * F \end{eqnarray}`

Next, rewrite the expression as `\blue{A}x^2 + \purple{a}x + \purple{b}x + \pink{C}`:

`\qquad \blue{A}x^2 +\purple{E}x +\purple{A * F}x +\pink{C}`

Group the terms so that there is a common factor in each group:

`\qquad (\blue{A}x^2 +\purple{E}x) + (\purple{A * F}x +\pink{C})`

Factor out the common factors:

`\qquad x(Ax + E) + F(Ax + E)`

Notice how `(Ax + E)` has become a common factor. Factor this out to find the answer.

`(Ax + E)(x + F)`