# How do you use synthetic division to divide x^4-4x^2+7x+15 divided by x+4?

Jan 29, 2016

${x}^{3} - 4 {x}^{2} + 12 x - 41$ with a remainder of 179.

#### Explanation:

${x}^{4} - 4 {x}^{2} + 7 x + 15$ divided by $x + 4$ gives ${x}^{3}$ together with a remainder, because the high order term of the divisor, namely x, times ${x}^{3}$ yields the high order term of the polynomial being divided, namely ${x}^{4}$. To find the remainder, multiply $x + 4$ times ${x}^{3}$ and get ${x}^{4} + 4 {x}^{3}$. Subtract that from ${x}^{4} - 4 {x}^{2} + 7 x + 15$ to get $- 4 {x}^{3} - 4 {x}^{2} + 7 x + 15$. That is the initial remainder.

Divide that again by x+4, which goes $- 4 {x}^{2}$ times, because the high order term of the divisor, namely x, times $- 4 {x}^{2}$, yields the high order term of this initial remainder, namely, $- 4 {x}^{3}$. To find the new remainder, multiply $x + 4$ times $- 4 {x}^{2}$ and get $- 4 {x}^{3} - 16 {x}^{2}$. Subtract that from the initial remainder, namely $- 4 {x}^{3} - 4 {x}^{2} + 7 x + 15$ and get the secondary remainder, namely $12 {x}^{2} + 7 x + 15$

Divide the secondary remainder by the divisor $x + 4$ which goes $12 x$ times. To get the third remainder multiply $12 x$ by $x + 4$ which yields $12 {x}^{2} + 48$. Subtract this from the secondary remainder, namely $12 {x}^{2} + 7 x + 15$, to get the third remainder, namely $41 x + 15$.

Divide the third remainder by $x + 4$ which goes 41 times. To get the final remainder multiply $41$ times $x + 4$ which yields $41 x + 164$. Subtracting that from the third remainder,
namely $41 x + 15$ yields 179. The successive divisors are:
${x}^{3}$, $- 4 {x}^{2}$, $12 x$, and $- 41$. Adding these together yields the polynomial ${x}^{3} - 4 {x}^{2} + 12 x - 41$ together with the final remainder $179$.