How do you long divide $\frac{x^{3} - 13 x - 12}{x - 4}$ $(x^3-13x -12)/ (x-4)$ ?

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How do you long divide $\frac{x^{3} - 13 x - 12}{x - 4}$ $(x^3-13x -12)/ (x-4)$ ?

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Answer 1 · 2017-05-31T06:16:55

The quotient is $= x^{2} + 4 x + 3$ $=x^2+4x+3$ and the remainder is $= 0$ $=0$

Explanation:

Let's perform the long division

$a a a a$ $color(white)(aaaa)$ $x - 4$ $x-4$ $∣$ $|$ $a a a a$ $color(white)(aaaa)$ $x^{3} + 0 x^{2} - 13 x - 12$ $x^3+0x^2-13x-12$ $a a a a$ $color(white)(aaaa)$ $∣$ $|$ $x^{2} + 4 x + 3$ $x^2+4x+3$

$a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaa)$ $x^{3} - 4 x^{2}$ $x^3-4x^2$

$a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaa)$ $0 + 4 x^{2} - 13 x$ $0+4x^2-13x$

$a a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaaa)$ $+ 4 x^{2} - 16 x$ $+4x^2-16x$

$a a a a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaaaaa)$ $+ 0 + 3 x - 12$ $+0+3x-12$

$a a a a a a a a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaaaaaaaaa)$ $+ 3 x - 12$ $+3x-12$

$a a a a a a a a a a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaaaaaaaaaaa)$ $+ 0 - 0$ $+0-0$

Therefore,

$\frac{x^{3} - 13 x - 12}{x - 4} = x^{2} + 4 x + 3$ $(x^3-13x-12)/(x-4)=x^2+4x+3$

Answer 2 · 2017-05-31T07:45:14

$x^{2} + 4 x + 3$ $x^2+4x+3$

Explanation:

$one way is to use the divisor as a factor in the numerator$ $"one way is to use the divisor as a factor in the numerator"$

$consider the numerator$ $"consider the numerator"$

$x^{2} (x - 4) + 4 x^{2} - 13 x - 12$ $color(red)(x^2)(x-4)color(magenta)(+4x^2)-13x-12$

$= x^{2} (x - 4) + 4 x (x - 4) + 16 x - 13 x - 12$ $=color(red)(x^2)(x-4)color(red)(+4x)(x-4)color(magenta)(+16x)-13x-12$

$= x^{2} (x - 4) + 4 x (x - 4) + 3 (x - 4) + 12 - 12$ $=color(red)(x^2)(x-4)color(red)(+4x)(x-4)color(red)(+3)(x-4)color(magenta)(+12)-12$

$= x^{2} (x - 4) + 4 x (x - 4) + 3 (x - 4)$ $=color(red)(x^2)(x-4)color(red)(+4x)(x-4)color(red)(+3)(x-4)$

$quotient = x^{2} + 4 x + 3, remainder = 0$ $"quotient "=color(red)(x^2+4x+3)," remainder "=0$

$\Rightarrow \frac{x^{3} - 13 x - 12}{x - 4} = x^{2} + 4 x + 3$ $rArr(x^3-13x-12)/(x-4)=x^2+4x+3$

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How do you long divide $\frac{x^{3} - 13 x - 12}{x - 4}$ $(x^3-13x -12)/ (x-4)$ ?

2 Answers

Explanation:

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How do you long divide $\frac{x^{3} - 13 x - 12}{x - 4}$ $(x^3-13x -12)/ (x-4)$ ?