Q16. - Algebra 1

Question

Find each quotient.

$\left(3x^{\mathrm{2}}-14x-3\right)÷\left(x-5\right)$

Step-by-Step Solution

Answer

$\mathbf{3}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{+}\frac{\mathbf{2}}{\mathbf{x}\mathbf{-}\mathbf{5}}$

Step 1. Divide.

Divide the first term of the dividend, $3x^2$, by the first term of the divisor, $x$.

$$\begin{gathered} x-53x3x^2-14x-3 \\ \frac{{\displaystyle -3x^2-15x}}{{\displaystyle 1x-3}} \end{gathered}$$

Step 2. Divide.

Divide the first term of the partial dividend, $1x$, by the first term of the divisor, $x$.

$$\begin{gathered} x-53x+13x^2-14x-3 \\ \frac{{\displaystyle -3x^2-15x}}{{\displaystyle 1x-3}} \\ \frac{-1x-5}{2} \end{gathered}$$

Step 3. Write the conclusion.

Therefore, $\left(3x^2-14x-3\right)÷\left(x-5\right)$ is $3x+1$ with a remainder of 2. The answer can be written as $3x+1+\frac{2}{x-5}$.