Question

1. $(7v^4 + 32v^3 - 67v^2 - 88v + 138) div (7v - 10)$

Explanation:

Step1: Divide the leading terms

Divide the leading term of the dividend \(7v^4\) by the leading term of the divisor \(7v\). So, \(\frac{7v^4}{7v}=v^3\). Multiply the divisor \(7v - 10\) by \(v^3\) to get \(7v^4-10v^3\). Subtract this from the dividend:

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Step2: Divide the new leading term

Divide the leading term of the new dividend \(42v^3\) by the leading term of the divisor \(7v\). So, \(\frac{42v^3}{7v} = 6v^2\). Multiply the divisor \(7v - 10\) by \(6v^2\) to get \(42v^3-60v^2\). Subtract this from the new dividend:

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Step3: Divide the new leading term

Divide the leading term of the new dividend \(-7v^2\) by the leading term of the divisor \(7v\). So, \(\frac{-7v^2}{7v}=-v\). Multiply the divisor \(7v - 10\) by \(-v\) to get \(-7v^2 + 10v\). Subtract this from the new dividend:

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Step4: Divide the new leading term

Divide the leading term of the new dividend \(-98v\) by the leading term of the divisor \(7v\). So, \(\frac{-98v}{7v}=-14\). Multiply the divisor \(7v - 10\) by \(-14\) to get \(-98v + 140\). Subtract this from the new dividend:

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Step5: Combine the results

The quotient is \(v^3+6v^2 - v - 14\) and the remainder is \(-2\). So, \(\frac{7v^4 + 32v^3 - 67v^2 - 88v + 138}{7v - 10}=v^3 + 6v^2 - v - 14-\frac{2}{7v - 10}\) (or we can write it as \(v^3+6v^2 - v - 14\) with a remainder of \(-2\)). Wait, there was a mistake in step 2 earlier (the initial hand - written step had an error). Let's redo step 2 correctly:

When we do the first subtraction:

Dividend: \(7v^4+32v^3 - 67v^2-88v + 138\)

Divisor: \(7v - 10\)

First term of quotient: \(\frac{7v^4}{7v}=v^3\)

Multiply divisor by \(v^3\): \(7v^4-10v^3\)

Subtract: \((7v^4 + 32v^3-67v^2 - 88v + 138)-(7v^4 - 10v^3)=7v^4+32v^3-67v^2 - 88v + 138 - 7v^4 + 10v^3 = 42v^3-67v^2 - 88v + 138\)

Second term of quotient: \(\frac{42v^3}{7v}=6v^2\)

Multiply divisor by \(6v^2\): \(42v^3-60v^2\)

Subtract: \((42v^3-67v^2 - 88v + 138)-(42v^3-60v^2)=42v^3-67v^2 - 88v + 138 - 42v^3 + 60v^2=-7v^2-88v + 138\) (this part was correct)

Third term of quotient: \(\frac{-7v^2}{7v}=-v\)

Multiply divisor by \(-v\): \(-7v^2 + 10v\)

Subtract: \((-7v^2-88v + 138)-(-7v^2 + 10v)=-7v^2-88v + 138 + 7v^2 - 10v=-98v + 138\) (correct)

Fourth term of quotient: \(\frac{-98v}{7v}=-14\)

Multiply divisor by \(-14\): \(-98v + 140\)

Subtract: \((-98v + 138)-(-98v + 140)=-98v + 138 + 98v - 140=-2\) (correct)

So the correct quotient is \(v^3 + 6v^2-v - 14\) and the remainder is \(-2\). So \((7v^4 + 32v^3 - 67v^2 - 88v + 138)\div(7v - 10)=v^3+6v^2 - v - 14+\frac{-2}{7v - 10}\) or \(v^3+6v^2 - v - 14\) with a remainder of \(-2\)

Answer:

The quotient is \(v^3 + 6v^2 - v - 14\) and the remainder is \(-2\), so \(\boldsymbol{(7v^4 + 32v^3 - 67v^2 - 88v + 138)\div(7v - 10)=v^3 + 6v^2 - v - 14-\frac{2}{7v - 10}}\) (or in quotient - remainder form: \(v^3+6v^2 - v - 14\) with remainder \(-2\))