# How do you use synthetic division to divide (x^4-11x^3+15x^2+40x-5) by x-5?

${x}^{3} - 6 {x}^{2} - 15 x - 35 + \frac{- 180}{x - 5}$

Remainder$= - 180$

#### Explanation:

The given: Dividend $\left({x}^{4} - 11 {x}^{3} + 15 {x}^{2} + 40 x - 5\right)$
and Divisor $\left(x - 5\right)$

Synthetic Division:
Arrange the terms from highest to lowest degree:

${x}^{4}$---- ${x}^{3}$----${x}^{2}$----${x}^{1}$----${x}^{0}$

Use the numerical coefficients only and the divisor $x - 5$ be equated to zero.

that is $x - 5 = 0$ solving for x: results to $x = + 5$ the trial divisor.

1 s t Line:$1 \setminus \setminus - 11 \setminus \setminus + 15 \setminus \setminus + 40 \setminus \setminus - 5$ use trial divisor=$+ 5$
2nd Line:$0 \setminus \setminus + 5 \setminus \setminus - 30 \setminus \setminus - 75 \setminus \setminus - 175$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
3rd Line:$1 \setminus \setminus - 6 \setminus \setminus - 15 \setminus \setminus - 35 \setminus \setminus - 180$

Bring down the first numerical coefficient 1. This number will be multiplied by the trial divisor 5 result is 5 which will be written at the 2nd column under $- 11$. Perform Algebraic addition using $- 11$ and $5$ and result is $- 6$ located at the second column of 3rd line.

Next $- 6$ be multiplied by the trial divisor $+ 5$, result is $- 30$ to be written at the 3rd column under the $+ 15$. Perform Algebraic addition using $+ 15$ and $- 30$ and result is $- 15$ located at the 3rd column of 3rd line.

Next $- 15$ be multiplied by the trial divisor $+ 5$, result is $- 75$ to be written at the 4th column under the $+ 40$. Perform Algebraic addition using $+ 40$ and $- 75$ and result is $- 35$ located at the 4th column of 3rd line.

Next $- 35$ be multiplied by the trial divisor $+ 5$, result is $- 175$ to be written at the 5th column under the $- 5$. Perform Algebraic addition using $- 5$ and $- 175$ and result is $- 180$ located at the 5th column of 3rd line. And this is the REMAINDER.

$\frac{D I V I D E N D}{D I V I S O R} = Q U O T I E N T + \frac{R E M A I N D E R}{D I V I S O R}$

$\frac{{x}^{4} - 11 {x}^{3} + 15 {x}^{2} + 40 x - 5}{x - 5} = {x}^{3} - 6 {x}^{2} - 15 x - 35 + \frac{- 180}{x - 5}$

Have a nice day!!! from the Philippines..