randRangeExclude(-9, 9, [-1, 0, 1]) randRangeNonZero(-9, 9) randRangeNonZero(-5, 5) E + A * F E * F
plus(A + "x", E) fractionReduce(-E, A, true) plus("x", F)

Simplify the right-hand side of the equation below and state the condition under which the simplification is valid.

`y = \dfrac{Ax^2+coefficient(B)x+C}{DENOMINATOR}`

`y =`
SOLUTION.toString()
`x \neq \space` -E/A
plus("x", F) -F plus(A+"x", E)

Simplify the right-hand side of the equation below and state the condition under which the simplification is valid.

`y = \dfrac{Ax^2+coefficient(B)x+C}{DENOMINATOR}`

`y =`
SOLUTION.toString()
`x \neq \space` -F

First use factoring by grouping to factor the expression in the numerator.

This expression is in the form `\blue{A}x^2 + \green{B}x + \pink{C}`.

$("#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)} &=& \red{A * C} \\ \purple{a} + \purple{b} &=& &=& \green{B} \end{eqnarray}`

In order to find `\purple{a}` and `\purple{b}`, list out the factors of `\red{A * C}` and add them together. Remember, since `\red{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}) &=& \red{A * C} \\ \purple{a} + \purple{b} &=& \purple{E} + \purple{A * F} &=& \green{B} \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})`

Factor out the common factors:

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

Now factor out `(Ax + E)`:

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

The original expression can therefore be written:

`\qquad y = \dfrac{(Ax + E)(x + F)}{DENOMINATOR}`

We can divide the numerator and denominator by `DENOMINATOR`:

`\qquad y = \dfrac{\cancel{(DENOMINATOR)}(SOLUTION)}{\cancel{DENOMINATOR}} = SOLUTION`

Because we divided by `DENOMINATOR`,

`\begin{align} DENOMINATOR &\neq 0 \\ x &\neq CONDITION \end{align}`

Therefore,

`\qquad y = SOLUTION`
`\qquad x \neq CONDITION`