# Khan/khan-exercises

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 Dividing polynomials by binomials 1
randVar() randVar() randRangeExclude(-9, 9, [-1, 0, 1]) -A * A

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

Y = \dfrac{X^2 + CONSTANT}{X + A}

Y =
X - A
\space X \neq \space -A

The numerator is in the form \pink{a^2} - \blue{b^2}, which is a difference of two squares so we can factor it as (\pink{a} + \blue{b})(\pink{a} - \blue{b}).

\qquad \blue{b = \sqrt{A * A} = abs(A)}

So we can rewrite the expression as:

Y = \dfrac{(\pink{X} + \blue{A})(\pink{X} \blue{-A})} \dfrac{(\pink{X} \blue{A})(\pink{X} + \blue{-A})} {X + A}

We can divide the numerator and denominator by (X + A):

Y = \dfrac{\cancel{(\pink{X} + \blue{A})}(\pink{X} \blue{-A})} \dfrac{\cancel{(\pink{X} \blue{A})}(\pink{X} + \blue{-A})} {\cancel{X + A}} = X + -A

Because we divided by (X + A),

\begin{align} X + A &\neq 0 \\ X &\neq -A \end{align}

Therefore,

\qquad Y = X - A

randVar() randVar() randRangeNonZero(-10, 10) randRangeNonZero(-10, 10) A * B -A - B

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

Y = \dfrac{X^2 + plus(LINEAR + X) + CONSTANT}{X - A}

Y =
X - B
\space X \neq \space A

X^2 + plus(LINEAR + X) + CONSTANT = (X - A)(X - B)

So we can rewrite the expression as:

Y = \dfrac{(X + -A)(X + -B)}{X + -A}

We can divide the numerator and denominator by (X - A):

Y = \dfrac{\cancel{(X - A)}(X - B)}{\cancel{X - A}} = X - B

Because we divided by (X - A),

\begin{align} X + -A &\neq 0 \\ X &\neq A \end{align}

Therefore,

\qquad Y = X - B