# Khan/khan-exercises

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 Dividing polynomials by binomials 3
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,