How do you divide $\frac{5 x^{3} - 7 x^{2} - 4 x + 1}{3 x - 1}$ $(5x^3-7x^2-4x+1)/(3x-1)$ ?

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How do you divide $\frac{5 x^{3} - 7 x^{2} - 4 x + 1}{3 x - 1}$ $(5x^3-7x^2-4x+1)/(3x-1)$ ?

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Answer 1 · 2018-02-04T14:04:07

Use long division.

$\frac{5}{3} x^{2} - \frac{16}{9} x - \frac{52}{27} - \frac{25}{27 (3 x - 1)}$ $5/3x^2 - 16/9x - 52/27 -25/(27(3x-1))$

Explanation:

$(3 x - 1)$ $(3x-1)$ will go into $(5 x^{3} - 7 x^{2} - 4 x + 1)$ $(5x^3-7x^2-4x+1)$ a total of $\frac{5}{3} x^{2}$ $5/3x^2$ times.

$\frac{5}{3} x^{2} (3 x - 1) = (5 x^{3} - \frac{5}{3} x^{2})$ $5/3x^2 (3x - 1) = (5x^3 - 5/3x^2)$

This will leave a remainder of:

$\frac{5}{3} x^{2}$ $" "" "" "" ""5/3x^2$
$----------------------------$ $" "" "" "" ""----------------------------"$
$3 x - 1 ∣ 5 x^{3} - 7 x^{2} - 4 x + 1$ $3x-1" "| 5x^3-7x^2-4x+1$
$- (5 x^{3} - \frac{5}{3} x^{2})$ $" "" "" "-(5x^3 - 5/3x^2)$
$--------------$ $" "" "" "" ""--------------"$
$- \frac{16}{3} x^{2}$ $" "" "" "" "" "" "-16/3x^2$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So now we're left with $- \frac{16}{3} x^{2} - 4 x + 1$ $-16/3x^2-4x+1$ .

$(3 x - 1)$ $(3x-1)$ will go into this $- \frac{16}{9} x$ $-16/9x$ times.

$- \frac{16}{9} x (3 x - 1) = (- \frac{16}{3} x^{2} + \frac{16}{9} x)$ $-16/9x(3x - 1) = (-16/3x^2 + 16/9x)$

This will leave a remainder of:

$\frac{5}{3} x^{2} - \frac{16}{9} x$ $" "" "" "" ""5/3x^2 - 16/9x$
$----------------------------$ $" "" "" "" ""----------------------------"$
$3 x - 1 ∣ 5 x^{3} - 7 x^{2} - 4 x + 1$ $3x-1" "| 5x^3-7x^2-4x+1$
$- (5 x^{3} - \frac{5}{3} x^{2})$ $" "" "" "-(5x^3 - 5/3x^2)$
$-----------------$ $" "" "" "" ""-----------------"$
$- \frac{16}{3} x^{2} - 4 x + 1$ $" "" "" "" "" "" "-16/3x^2 -4x+1$
$- (- \frac{16}{3} x^{2} + \frac{16}{9} x)$ $" "" "" "" "-(-16/3x^2+16/9x)$
$-------------------------$ $" "" "" "" "" ""-------------------------"$
$- \frac{52}{9} x$ $" "" "" "" "" "" "" "" "" "-52/9x$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So now we're left with $- \frac{52}{9} x + 1$ $-52/9x + 1$ .

$(3 x - 1)$ $(3x-1)$ will go into this $- \frac{52}{27}$ $-52/27$ times.

$- \frac{52}{27} (3 x - 1) = (- \frac{52}{9} x + \frac{52}{27})$ $-52/27(3x-1) = (-52/9x + 52/27)$

This will leave a remainder of:

$\frac{5}{3} x^{2} - \frac{16}{9} x - \frac{52}{27}$ $" "" "" "" ""5/3x^2 - 16/9x - 52/27$
$----------------------------$ $" "" "" "" ""----------------------------"$
$3 x - 1 ∣ 5 x^{3} - 7 x^{2} - 4 x + 1$ $3x-1" "| 5x^3-7x^2-4x+1$
$- (5 x^{3} - \frac{5}{3} x^{2})$ $" "" "" "-(5x^3 - 5/3x^2)$
$-----------------$ $" "" "" "" ""-----------------"$
$- \frac{16}{3} x^{2} - 4 x + 1$ $" "" "" "" "" "" "-16/3x^2 -4x+1$
$- (- \frac{16}{3} x^{2} + \frac{16}{9} x)$ $" "" "" "" "-(-16/3x^2+16/9x)$
$-------------------------$ $" "" "" "" "" ""-------------------------"$
$- \frac{52}{9} x + 1$ $" "" "" "" "" "" "" "" "" "-52/9x + 1$
$- (- \frac{52}{9} x + \frac{52}{27})$ $" "" "" "" "" "" "" "-(-52/9x + 52/27)$
$-------------------------$ $" "" "" "" "" "" "" ""-------------------------"$
$MMMMMMMMMMMMMM- - \frac{25}{27}$ $color(white)"MMMMMMMMMMMMMM-"-25/27$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Since the degree of this remainder is smaller than the degree of our divisor, we will just divide it by the divisor as the last term in our answer:

$\frac{5}{3} x^{2} - \frac{16}{9} x - \frac{52}{27} - \frac{\frac{25}{27}}{3 x - 1}$ $5/3x^2 - 16/9x - 52/27 -(25/27)/(3x-1)$

$\frac{5}{3} x^{2} - \frac{16}{9} x - \frac{52}{27} - \frac{25}{27 (3 x - 1)}$ $5/3x^2 - 16/9x - 52/27 -25/(27(3x-1))$

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Explanation:

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