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drag the tiles to the correct boxes to complete the pairs. not all tile…

Question

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.
division expressions:
\\(\frac{-66x^2 + 1633x - 77}{11x - 7}\\)
\\(\frac{-55x^2 + 68x - 21}{-5x + 3}\\)
\\(\frac{-14x^2 - 68x - 48}{2x + 8}\\)
tiles (quotients):
-7x - 6
-6x + 11
11x - 7
6x + 11
7x - 6
7x + 6
11x + 7
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Explanation:

Step 1: Divide \(\frac{-14x^{2}-68x - 48}{2x + 8}\)

First, factor out \(-2\) from the numerator: \(-2(7x^{2}+34x + 24)\). Then factor the quadratic in the numerator: \(7x^{2}+34x + 24=(7x + 6)(x + 4)\). The denominator is \(2x + 8 = 2(x + 4)\). Cancel out \((x + 4)\) and divide the remaining terms: \(\frac{-2(7x + 6)(x + 4)}{2(x + 4)}=-7x - 6\).

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

Use polynomial long division or factor. Let's try factoring. The numerator: \(-55x^{2}+68x - 21 =-(55x^{2}-68x + 21)\). Factor \(55x^{2}-68x + 21=(11x - 7)(5x - 3)\). So the expression becomes \(\frac{-(11x - 7)(5x - 3)}{-(5x - 3)} = 11x - 7\) (note: original denominator is \(-5x + 3=-(5x - 3)\)).

Step 3: Divide \(\frac{-66x^{2}+163x - 77}{11x - 7}\)

Factor the numerator: \(-66x^{2}+163x - 77=-(66x^{2}-163x + 77)\). Factor \(66x^{2}-163x + 77=(6x - 11)(11x - 7)\) (wait, let's check: \((6x - 11)(11x - 7)=66x^{2}-42x - 121x + 77=66x^{2}-163x + 77\)). So the expression is \(\frac{-(6x - 11)(11x - 7)}{11x - 7}=-6x + 11\).

Answer:

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