Problem 86 Divide. Tell whether each diviso... [FREE SOLUTION]

Chapter 7: Problem 86

Divide. Tell whether each divisor is a factor of the dividend. $$ \left(6 a^{3}+a^{2}-a+4\right) \div(2 a+1) $$

Short Answer

Expert verified

The quotient is $3a^2 + a + 2$ and the remainder is 0. Thus, $2a + 1$ is indeed a factor of $6a^3 + a^2 - a + 4$.

Step by step solution

Set-up the Long Division

To start, use long division for polynomials. Write $6a^3 + a^2 - a + 4$ inside the division symbol and $2a + 1$ outside, similar to how regular long division is set up.

Start Division

Divide the first term of the dividend ($6a^3$) by the first term in the divisor ($2a$). The result is $3a^2$. This is your first term of your quotient. You now multiply $2a+1$ by $3a^2$ and subtract the result from the dividend.

Continue Division

The subtraction result is $2a^2 - a + 4$. Copy the new polynomial below and repeat the division process again. Divide $2a^2$ by $2a$ gives $a$, which is then subtracted from the new dividend. Hence we obtain the new quotient as $3a^2 + a$.

Continue until Remainder

Continue the process until you are left with a constant term or a degree lower than the divisor. The final quotient is $3a^2 + a + 2$, and the remainder is $0$.

Factor Determination

Considering there is no remainder, we conclude that the divisor is a factor of the dividend. In other words, $2a + 1$ is a factor of $6a^3 + a^2 - a + 4$.

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91影视

Divide. Tell whether each divisor is a factor of the dividend. $$ \left(6 a^{3}+a^{2}-a+4\right) \div(2 a+1) $$

Short Answer

Step by step solution

Set-up the Long Division

Start Division

Continue Division

Continue until Remainder

Factor Determination

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