# Khan/khan-exercises

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 Factoring polynomials by grouping
randRangeNonZero(-9, 9) randRangeNonZero(-9, 9) randRangeNonZero(-5, 5) E + A * F E * F

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

Ax^2+ Bx+ 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)