Algebraic Division
Question
Use long division to divide \((5x^4 + 12x^3 - 25x^2 + 75)\) by \((x + 3)\)
Solution
Show solution Hide solution Fully worked — 5 steps
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Find the first term of the quotient
Divide \(5x^4\) by \(x\) to get the first term of the quotient, \(5x^3\). Multiply the divisor \((x + 3)\) by \(5x^3\) and subtract:
\[ 5x^4 \div x = 5x^3 \]\[ (5x^3)(x + 3) = 5x^4 + 15x^3 \]\[ (5x^4 + 12x^3) - (5x^4 + 15x^3) = -3x^3 \]Bring down the next term, \(-25x^2\); the working line is now \(-3x^3 - 25x^2\).
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Find the second term of the quotient
Divide \(-3x^3\) by \(x\) to get the second term of the quotient, \(-3x^2\). Multiply the divisor \((x + 3)\) by \(-3x^2\) and subtract:
\[ -3x^3 \div x = -3x^2 \]\[ (-3x^2)(x + 3) = -3x^3 - 9x^2 \]\[ (-3x^3 - 25x^2) - (-3x^3 - 9x^2) = -16x^2 \]Bring down the next term, \(0\); the working line is now \(-16x^2 + 0\).
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Find the third term of the quotient
Divide \(-16x^2\) by \(x\) to get the third term of the quotient, \(-16x\). Multiply the divisor \((x + 3)\) by \(-16x\) and subtract:
\[ -16x^2 \div x = -16x \]\[ (-16x)(x + 3) = -16x^2 - 48x \]\[ (-16x^2 + 0) - (-16x^2 - 48x) = 48x \]Bring down the next term, \(75\); the working line is now \(48x + 75\).
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Find the fourth term of the quotient
Divide \(48x\) by \(x\) to get the fourth term of the quotient, \(48\). Multiply the divisor \((x + 3)\) by \(48\) and subtract:
\[ 48x \div x = 48 \]\[ (48)(x + 3) = 48x + 144 \]\[ (48x + 75) - (48x + 144) = -69 \] -
State the result
In this case the remainder is \(-69\). Therefore:
\[ (5x^4 + 12x^3 - 25x^2 + 0 + 75) \div (x + 3) = 5x^3 - 3x^2 - 16x + 48 - \frac{69}{x + 3} \]That is, the quotient is \(5x^3 - 3x^2 - 16x + 48\) and the remainder is \(-69\).