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bio_man bio_man
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Posts: 31045
3 months ago
Divide by synthetic division: \((2x^3+x^2-22x+20)\div (2x-3)\)



Begin by solving for \(x\) in the divisor:

\(2x-3=0\)

\(\displaystyle x=\frac{3}{2}\)

Now perform synthetic division:



Given a remainder of \(4\), write the expression as:

\(\displaystyle 2x^2+4x-16-\frac{4}{x-\frac{3}{2}}\)

Recall that you set the divisor \(2x-3\) equal to \(0\) at the start. Since \(x\) has a coefficient of \(2\), divide the expression above by \(2\) as well.

\(\displaystyle \frac{\left(2x^2+4x-16-\frac{4}{x-\frac{3}{2}}\right)}{2}\)

Simplify by finding a common denominator in the numerator:

\(\displaystyle =\frac{\frac{\left(2x^2+4x-16\right)\left(x-\frac{3}{2}\right)-4}{x-\frac{3}{2}}}{2}\)

\(\displaystyle =\frac{\left(2x^2+4x-16\right)\left(x-\frac{3}{2}\right)-4}{x-\frac{3}{2}}\div 2\)

\(\displaystyle =\frac{\left(2x^2+4x-16\right)\left(x-\frac{3}{2}\right)-4}{x-\frac{3}{2}}\times \frac{1}{2}\)

Expand and multiply:

\(\displaystyle = \frac{2x^3+x^2-22x+22}{2x-3} \)
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