Problem 86 - Algebra 2

Question

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

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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.

Step 3: 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 \(3a^2 + a\).

Step 4: 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 \(0\).

Step 5: 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 \(6a^3 + a^2 - a + 4\).