Question

Use synthetic division with the factor x + 1 to completely factor LaTeX: x^3+2x^2-5x-6x 3 + 2 x 2 − 5 x − 6.

Answer 1

Answer:

$x^3+2x^2-5x-6=\left(x+1\right)\left(x-2\right)\left(x+3\right)$.

Step-by-step explanation:

To find $\frac{x^{3} + 2 x^{2} - 5 x - 6}{x + 1}$ using synthetic division you must:

Write the problem in a division-like format. To do this:

Take the constant term of the divisor with the opposite sign and write it to the left.

Write the coefficients of the dividend to the right.

$\begin{array}{c|cccc}&x^{3}&x^{2}&x^{1}&x^{0}\\-1&1&2&-5&-6\\&&\\\hline&\end{array}$

Step 1: Write down the first coefficient without changes:

$\begin{array}{c|rrrr}-1&1&2&-5&-6\\&&\\\hline&\1\end{array}$

Step 2:

Multiply the entry in the left part of the table by the last entry in the result row.

Add the obtained result to the next coefficient of the dividend, and write down the sum.

$\begin{array}{c|rrrr}-1&1&2&-5&-6\\&&\left(-1\right) \cdot 1=-1\\\hline&{1}&{2}+\left({-1}\right)={1}\end{array}$

Step 3:

Multiply the entry in the left part of the table by the last entry in the result row.

Add the obtained result to the next coefficient of the dividend, and write down the sum.

$\begin{array}{c|rrrr}{-1}&1&2&{-5}&-6\\&&-1&\left({-1}\right) \cdot {1}={-1}\\\hline&1&{1}&\left({-5}\right)+\left({-1}\right)={-6}\end{array}$

Step 4:

Multiply the entry in the left part of the table by the last entry in the result row.

Add the obtained result to the next coefficient of the dividend, and write down the sum.

$\begin{array}{c|rrrr}{-1}&1&2&-5&{-6}\\&&-1&-1&\left({-1}\right) \cdot \left({-6}\right)={6}\\\hline&1&1&{-6}&\left({-6}\right)+{6}={0}\end{array}$

We have completed the table and have obtained the following resulting coefficients: 1, 1, −6, 0.

All the coefficients except the last one are the coefficients of the quotient, the last coefficient is the remainder.

Thus, the quotient is $x^{2}+x-6$, and the remainder is 0.

$\frac{x^{3} + 2 x^{2} - 5 x - 6}{x + 1}=x^{2} + x - 6+\frac{0}{x + 1}=x^{2} + x - 6$

Now, we factor $x^{2}+x-6$

$\left(x^2-2x\right)+\left(3x-6\right)\\x\left(x-2\right)+3\left(x-2\right)\\\left(x-2\right)\left(x+3\right)$

Therefore,

$x^3+2x^2-5x-6=\left(x+1\right)\left(x-2\right)\left(x+3\right)$