Question

5
drag the tiles to the correct boxes to complete the pairs. not all tiles will be used.
match each division expression with the correct quotient.
11x - 7 11x + 7 7x + 6 -7x - 6 7x - 6
6x + 11 -6x + 11
\\(\frac{-66x^2 + 163x - 77}{11x - 7}\\) →
\\(\frac{-55x^2 + 68x - 21}{-5x + 3}\\) →
\\(\frac{-14x^2 - 68x - 48}{2x + 8}\\) →

Explanation:

Step1: Solve \(\frac{-66x^2 + 163x - 77}{11x - 7}\)

We can use polynomial long division or factor the numerator. Let's try factoring. We assume the numerator factors as \((ax + b)(11x - 7)\). Expanding \((ax + b)(11x - 7)=11ax^2+(11b - 7a)x - 7b\). Comparing with \(-66x^2 + 163x - 77\), we have \(11a=-66\Rightarrow a = - 6\). Then \(11b-7(-6)=163\Rightarrow11b + 42 = 163\Rightarrow11b=121\Rightarrow b = 11\). So the numerator is \((-6x + 11)(11x - 7)\). Then \(\frac{(-6x + 11)(11x - 7)}{11x - 7}=-6x + 11\) (canceling \(11x - 7\) as \(11x-7
eq0\)).

Step2: Solve \(\frac{-55x^2 + 68x - 21}{-5x + 3}\)

Factor the numerator. Assume \((ax + b)(-5x + 3)=-5ax^2+(3a - 5b)x + 3b\). Comparing with \(-55x^2 + 68x - 21\), we have \(-5a=-55\Rightarrow a = 11\). Then \(3(11)-5b=68\Rightarrow33 - 5b=68\Rightarrow - 5b=35\Rightarrow b=-7\). So numerator is \((11x - 7)(-5x + 3)\). Then \(\frac{(11x - 7)(-5x + 3)}{-5x + 3}=11x - 7\) (canceling \(-5x + 3\) as \(-5x + 3
eq0\)). Wait, no, wait: Wait, \((11x - 7)(-5x + 3)=-55x^2+33x + 35x - 21=-55x^2 + 68x - 21\), correct. So \(\frac{(11x - 7)(-5x + 3)}{-5x + 3}=11x - 7\)? Wait, no, wait the denominator is \(-5x + 3\), and the numerator factors to \((11x - 7)(-5x + 3)\), so cancel \(-5x + 3\) (assuming \(-5x + 3
eq0\)) to get \(11x - 7\)? Wait, no, wait \((11x - 7)(-5x + 3)=-55x^2+33x + 35x - 21=-55x^2 + 68x - 21\), yes. So \(\frac{(11x - 7)(-5x + 3)}{-5x + 3}=11x - 7\) (when \(-5x+3
eq0\)).

Wait, maybe I made a mistake. Let's do polynomial long division for \(\frac{-55x^2 + 68x - 21}{-5x + 3}\). Divide \(-55x^2\) by \(-5x\) to get \(11x\). Multiply \(-5x + 3\) by \(11x\) to get \(-55x^2 + 33x\). Subtract from the numerator: \((-55x^2 + 68x - 21)-(-55x^2 + 33x)=35x - 21\). Now divide \(35x\) by \(-5x\) to get \(-7\). Multiply \(-5x + 3\) by \(-7\) to get \(35x - 21\). Subtract: \((35x - 21)-(35x - 21)=0\). So the quotient is \(11x - 7\).

Step3: Solve \(\frac{-14x^2 - 68x - 48}{2x + 8}\)

First, factor numerator and denominator. Denominator: \(2x + 8 = 2(x + 4)\). Numerator: factor out \(-2\): \(-2(7x^2 + 34x + 24)\). Now factor \(7x^2 + 34x + 24\). We need two numbers \(m\) and \(n\) such that \(m\times n=7\times24 = 168\) and \(m + n = 34\). The numbers are 28 and 6. So \(7x^2+28x + 6x + 24=7x(x + 4)+6(x + 4)=(7x + 6)(x + 4)\). So numerator is \(-2(7x + 6)(x + 4)\). Denominator is \(2(x + 4)\). Cancel \(2(x + 4)\) (assuming \(x
eq - 4\)): \(\frac{-2(7x + 6)(x + 4)}{2(x + 4)}=- (7x + 6)=-7x - 6\).

Answer:

• \(\frac{-66x^2 + 163x - 77}{11x - 7}\) pairs with \(-6x + 11\)
• \(\frac{-55x^2 + 68x - 21}{-5x + 3}\) pairs with \(11x - 7\)
• \(\frac{-14x^2 - 68x - 48}{2x + 8}\) pairs with \(-7x - 6\)