How do you divide $(x^{3} + 8 x^{2} - 3 x + 16) \div (x^{2} + 5)$ $(x^3+8x^2-3x+16)div(x^2+5)$ using long division?

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Answer 1 · 2018-01-12T10:27:43

The quotient is $= (x + 8)$ $=(x+8)$ and the remainder is $= - (8 x + 24)$ $=-(8x+24)$

Perform the long division

$a a a a$ $color(white)(aaaa)$ $x^{3} + 8 x^{2} - 3 x + 16$ $x^3+8x^2-3x+16$ $a a a a$ $color(white)(aaaa)$ $∣$ $|$ $x^{2} + 5$ $x^2+5$

$a a a a$ $color(white)(aaaa)$ $x^{3} + 0 x^{2} + 5 x$ $x^3+0x^2+5x$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $∣$ $|$ $x + 8$ $x+8$

$a a a a a$ $color(white)(aaaaa)$ $0 + 8 x^{2} - 8 x + 16$ $0+8x^2-8x+16$

$a a a a a a a$ $color(white)(aaaaaaa)$ $+ 8 x^{2} + 0 x + 40$ $+8x^2+0x+40$

$a a a a a a a a a$ $color(white)(aaaaaaaaa)$ $+ 0 - 8 x - 24$ $+0-8x-24$

Therefore,

$\frac{x^{3} + 8 x^{2} - 3 x + 16}{x^{2} + 5} = (x + 8) - \frac{8 x + 24}{x^{2} + 5}$ $(x^3+8x^2-3x+16)/(x^2+5)=(x+8)-(8x+24)/(x^2+5)$

The quotient is $= (x + 8)$ $=(x+8)$ and the remainder is $= - (8 x + 24)$ $=-(8x+24)$

How do you divide (x^3+8x^2-3x+16)div(x^2+5)(x3+8x2−3x+16)÷(x2+5)(x^3+8x^2-3x+16)div(x^2+5) using long division?