How do you use synthetic division to divide $x^{3} + 2 x^{2} - 2 x + 24$ $x^3 + 2x^2 - 2x + 24$ by $x + 4$ $x+4$ ?

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How do you use synthetic division to divide $x^{3} + 2 x^{2} - 2 x + 24$ $x^3 + 2x^2 - 2x + 24$ by $x + 4$ $x+4$ ?

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Answer 1 · 2018-05-18T16:42:59

$\frac{x^{3} + 2 x^{2} - 2 x + 24}{x + 4} = x^{2} - 2 x + 6$ $frac{x^3 + 2x^2 - 2x + 24}{x + 4} = x^2 - 2x + 6$

Explanation:

Divide $x^{3}$ $x^3$ by $x + 4$ $x + 4$
Quotient $= x^{2}$ $= x^2$
Remainder $= x^{3} + 2 x^{2} - 2 x + 24 - x^{3} - 4 x^{2}$ $= x^3 + 2x^2 - 2x + 24 - x^3 - 4 x ^2$
Remainder $= - 2 x^{2} - 2 x + 24$ $= - 2x^2 - 2x + 24$

Divide $- 2 x^{2}$ $-2x^2$ by $x + 4$ $x + 4$
Quotient $= - 2 x$ $= -2x$
Remainder $= - 2 x^{2} - 2 x + 24 + 2 x^{2} + 8 x$ $= - 2x^2 - 2x + 24 + 2x^2 + 8x$
Remainder $= 6 x + 24$ $= 6x + 24$

Divide $6 x$ $6x$ by $x + 4$ $x + 4$
Quotient $= 6$ $=6$
Remainder $= 0$ $= 0$

Answer 2 · 2018-05-18T19:53:34

Synthetic division gives: $x^{2} - 2 x + 6$ $x^2-2x+6$

Process shown in detail

Explanation:

Given: $\frac{x^{3} + 2 x^{2} - 2 x + 24}{x + 4}$ $(x^3+2x^2-2x+24)/(x+4)$

Consider the denominator $x + 4$ $x+4$

Set $x + 4 = 0 \Rightarrow x = - 4$ $x+4=0 =>color(red)(x=-4)$

Now we construct the coefficient manipulation.

$dddd x^{3} + 2 x^{2} - 2 x + 24$ $color(white)("dddd") x^3+2x^2-2x+24$
$ddd ↓ d.d ↓ ddd ↓ ddd ↓$ $color(white)("ddd")darrcolor(white)("d.d") darrcolor(white)("ddd")darrcolor(white)("ddd")darr$

$- 4 ∣ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 1 ddd 2 ddd - 2 ddd 24$ $color(red)(-4)|bar(1color(white)("ddd")2color(white)("ddd")-2color(white)("ddd")24$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
bring down the first value of $1$ $color(magenta)(1)$

$- 4 ∣ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 1 ddd 2 ddd - 2 ddd 24$ $color(red)(-4)|bar(1color(white)("ddd")2color(white)("ddd")-2color(white)("ddd")24$
$color(white)("d..d")ul(|color(magenta)(darr )$
$color(white)("dddd")color(magenta)(1)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$-4xxcolor(magenta)(1)=color(blue)(-4)$
Place the $color(blue)(-4)$ under the 2

$color(red)(-4)|bar(1color(white)("ddd")2color(white)("ddd")-2color(white)("ddd")24$
$color(white)("d..d")ul(|color(magenta)(darr )color(white)("d")color(blue)(-4) )$
$color(white)("dddd")color(magenta)(1)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Add the 2 and the $color(blue)(-4)$

$color(red)(-4)|bar(1color(white)("ddd")2color(white)("ddd")-2color(white)("ddd")24$
$color(white)("d..d")ul(|color(magenta)(darr )color(white)("d")color(blue)(-4) )$
$color(white)("dddd")color(magenta)(1)color(white)("d")-2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(red)(-4)xx" the new "-2=+8$
Place the $-8$ under the $-2$

$color(red)(-4)|bar(1color(white)("dddd")2color(white)("ddd")-2color(white)("ddd")24$
$color(white)("d..d")ul(|darr color(white)("d")-4color(white)("ddddd")8 )$
$color(white)("dddd")1color(white)("dd")-2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Add the $-2" and the "8$

$color(red)(-4)|bar(1color(white)("dddd")2color(white)("ddd")-2color(white)("ddd")24$
$color(white)("d..d")ul(|darr color(white)("d")-4color(white)("ddddd")8 )$
$color(white)("dddd")1color(white)("dd")-2color(white)("ddddd")6$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(red)(-4)xx "the new "6 = -24$
Place the -24 under the existing +24
Add and you get the 0 so it is an exact division. No remainder.

$color(red)(-4)|bar(1color(white)("dddd")2color(white)("ddd")-2color(white)("ddd")24$
$color(white)("d..d")ul(|darr color(white)("d")-4color(white)("ddddd")8color(white)("ddd")24 )$
$color(white)("dddd")1color(white)("dd")-2color(white)("ddddd")6color(white)("ddd")0$
$color(white)("ddd.")darrcolor(white)("ddd")darrcolor(white)("dddd")darr$

$color(white)("dddd")x^2color(white)("d")-2xcolor(white)("dd")+6$

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How do you use synthetic division to divide $x^{3} + 2 x^{2} - 2 x + 24$ $x^3 + 2x^2 - 2x + 24$ by $x + 4$ $x+4$ ?

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How do you use synthetic division to divide $x^{3} + 2 x^{2} - 2 x + 24$ $x^3 + 2x^2 - 2x + 24$ by $x + 4$ $x+4$ ?