How do you divide $\frac{3 x^{4} + 2 x^{3} - 11 x^{2} - 2 x + 5}{x^{2} - 2}$ $(3x^4 + 2x^3 - 11x^2 - 2x + 5)/ (x^2 - 2)$ using polynomial long division?

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How do you divide $\frac{3 x^{4} + 2 x^{3} - 11 x^{2} - 2 x + 5}{x^{2} - 2}$ $(3x^4 + 2x^3 - 11x^2 - 2x + 5)/ (x^2 - 2)$ using polynomial long division?

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Answer 1 · 2017-11-14T20:29:45

$\frac{3 x^{4} + 2 x^{3} - 11 x^{2} - 2 x + 5}{x^{2} - 2} = 3 x^{2} + 2 x - 5 + \frac{2 x - 5}{x^{2} - 2}$ $(3x^4+2x^3-11x^2-2x+5)/(x^2-2)=3x^2+2x-5+(2x-5)/(x^2-2)$

Explanation:

$3 x^{4} + 2 x^{3} - 11 x^{2} - 2 x + 5 ∣ x^{2} - 2$ $3x^4+2x^3-11x^2-2x+5|x^2-2$
you ask what is it $\frac{3 x^{4}}{x^{2}}$ $(3x^4)/x^2$ and you get $3 x^{2}$ $3x^2$

(REMEMBER $3 x^{2}$ $3x^2$ )

now yow duplicate $3 x^{2}$ $3x^2$ to $x^{2} - 2$ $x^2-2$ and get $3 x^{4} - 6 x^{2}$ $3x^4-6x^2$
and do:
$3 x^{4} + 2 x^{3} - 11 x^{2} - 2 x + 5$ $3x^4+2x^3-11x^2-2x+5$
$-$ $-$
$3 x^{4} + 0 x^{3} - 6 x^{2} + 0 x + 0$ $3x^4+0x^3-6x^2+0x+0$
$=$ $=$
$0 + 2 x^{3} - 5 x^{2} - 2 x + 5$ $0+2x^3-5x^2-2x+5$
so we have now $2 x^{3} - 5 x^{2} - 2 x + 5$ $2x^3-5x^2-2x+5$ which is very exiting because we don't have the power of 4 anymore!

now we do the same again, try to follow:

First step (write it down):
$2 x^{3} - 5 x^{2} - 2 x + 5 ∣ x^{2} - 2$ $2x^3-5x^2-2x+5|x^2-2$

Second step (divided strongest power):
$\frac{2 x^{3}}{x^{2}} = 2 x$ $(2x^3)/(x^2)=2x$

(REMEMBER $2 x$ $2x$ )

Third step (Second step times the divider):
$2 x \cdot (x^{2} - 2) = 2 x^{3} - 4 x$ $2x*(x^2-2)=2x^3-4x$

Forth step (which is first step minus third step):
$2 x^{3} - 5 x^{2} - 2 x + 5$ $2x^3-5x^2-2x+5$
$-$ $-$
$2 x^{3} + 0 x^{2} - 4 x + 0$ $2x^3+0x^2-4x+0$
$=$ $=$
$0 - 5 x^{2} + 2 x + 5$ $0-5x^2+2x+5$

--and agian--

First step (write it down):
$- 5 x^{2} + 2 x + 5 ∣ x^{2} - 2$ $-5x^2+2x+5|x^2-2$

Second step (divided strongest power):
$\frac{- 5 x^{2}}{x^{2}} = - 5$ $(-5x^2)/(x^2)=-5$

(REMEMBER $- 5$ $-5$ )

Third step (Second step times the divider):
$- 5 \cdot (x^{2} - 2) = - 5 x^{2} + 10$ $-5*(x^2-2)=-5x^2+10$

Forth step (which is first step minus third step):
$- 5 x^{2} + 2 x + 5$ $-5x^2+2x+5$
$-$ $-$
$- 5 x^{2} + 10$ $-5x^2+10$
$=$ $=$
$0 + 2 x - 5$ $0+2x-5$

All REMEMBERs are the result:
$3 x^{2} + 2 x - 5$ $3x^2+2x-5$

BUT in that case, $2 x - 5$ $2x-5$ could not be divided so it remains as $\frac{2 x - 5}{x^{2} - 2}$ $(2x-5)/(x^2-2)$

SOOOOOOOOOOOOOOOOOOOOOOO :)

$\frac{3 x^{4} + 2 x^{3} - 11 x^{2} - 2 x + 5}{x^{2} - 2} = 3 x^{2} + 2 x - 5 + \frac{2 x - 5}{x^{2} - 2}$ $(3x^4+2x^3-11x^2-2x+5)/(x^2-2)=3x^2+2x-5+(2x-5)/(x^2-2)$

Lets check:

$(3 x^{2} + 2 x - 5 + \frac{2 x - 5}{x^{2} - 2}) (x^{2} - 2) =$ $(3x^2+2x-5+(2x-5)/(x^2-2))(x^2-2)=$
$= 3 x^{4} - 6 x^{2} + 2 x^{3} - 4 x - 5 x^{2} + 10 + 2 x - 5 =$ $=3x^4-6x^2+2x^3-4x-5x^2+10+2x-5=$
$= 3 x^{4} + 2 x^{3} - 11 x^{2} - 2 x + 5$ $=3x^4+2x^3-11x^2-2x+5$

So it's fine :)

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How do you divide $\frac{3 x^{4} + 2 x^{3} - 11 x^{2} - 2 x + 5}{x^{2} - 2}$ $(3x^4 + 2x^3 - 11x^2 - 2x + 5)/ (x^2 - 2)$ using polynomial long division?

1 Answer

Explanation:

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Explanation:

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How do you divide $\frac{3 x^{4} + 2 x^{3} - 11 x^{2} - 2 x + 5}{x^{2} - 2}$ $(3x^4 + 2x^3 - 11x^2 - 2x + 5)/ (x^2 - 2)$ using polynomial long division?