How do you use synthetic division to divide $(x^{2} + 13 x + 40) \div (x + 8)$ $(x^2 + 13x + 40) ÷ (x + 8)$ ?

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Answer 1 · 2015-05-16T21:25:23

Doing synthetic division is rather like doing long division.

First look for a multiplier for $(x + 8)$ $(x+8)$ that will match the highest term $x^{2}$ $x^2$ . That multiplier must be $x$ $x$ :

$x (x + 8) = x^{2} + 8 x$ $x(x+8) = x^2+8x$

Subtract the right hand side from the original $x^{2} + 13 x + 40$ $x^2+13x+40$ to find the remainder:

$(x^{2} + 13 x + 40) - (x^{2} + 8 x) = 5 x + 40$ $(x^2+13x+40)-(x^2+8x) = 5x+40$

Now look for a multiplier for $(x + 8)$ $(x+8)$ that will match the highest remaining term $5 x$ $5x$ . That multiplier must be $5$ $5$ :

$5 (x + 8) = 5 x + 40$ $5(x+8) = 5x+40$

Subtract the right hand side from our last remainder $5 x + 40$ $5x+40$ to find the remainder:

$(5 x + 40) - (5 x + 40) = 0$ $(5x+40)-(5x+40)=0$

Bingo! It divides perfectly.

Adding together the multipliers $x$ $x$ and $5$ $5$ that we found we get

$(x^{2} + 13 x + 40) \div (x + 8) = x + 5$ $(x^2+13x+40)-:(x+8) = x+5$

How do you use synthetic division to divide (x2+13x+40)÷(x+8)(x^2 + 13x + 40) ÷ (x + 8)?