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Chapter 2: Problem 23

Use long division to divide. \(\left(3 x+2 x^{3}-9-8 x^{2}\right) \div\left(x^{2}+1\right)\)

Short Answer

Expert verified
The quotient of dividing \(3x + 2x^3 - 9 - 8x^2\) by \(x^2 + 1\) is \(2x - 8\) and the remainder is \(9x - 1\).

Step by step solution

01

Arrange the Polynomial in Descending Order

Rearrange both the dividend and the divisor in descending order of powers. The dividend becomes \(2x^3 - 8x^2 + 3x - 9\) and the divisor remains \(x^2 + 1\) as it is already in the correct order.
02

Divide the Highest Degree Term of Dividend by Highest Degree Term of Divisor

Divide the highest degree term of the dividend (which is \(2x^3\)) by the highest degree term of the divisor (which is \(x^2\)). The result is \(2x\). This is the first term of our quotient.
03

Multiply the Divisor by the Quotient Obtained

Now, multiply \(x^2 + 1\) by \(2x\) resulting in \(2x^3 + 2x\).
04

Subtract this Result from Dividend

Subtract \(2x^3 + 2x\) from the rearranged dividend \(2x^3 - 8x^2 + 3x - 9\) to obtain \(-8x^2 + x - 9\) as the new dividend.
05

Repeat the Long Division Process

Now, repeat the process using \(-8x^2 + x - 9\) as the new dividend. The highest degree term is \(-8x^2\), which divided by \(x^2\) gives \(-8\). This is the next term in the quotient, so the quotient is now \(2x - 8\).
06

Continue until Degree of Remainder is Less than Degree of Divisor

Now multiply \(-8\) by the divisor \(x^2 + 1\) resulting in \(-8x^2 - 8\) and subtract it from the new dividend. This gives a remainder of \(9x - 1\). Since this remainder's highest degree term's degree (which is 1) is lower than the divisor's (which is 2), we stop the process here.

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