How do you divide $\frac{4 x^{3} + 3 x^{2} + 8 x + 12}{3 x - 4}$ $(4x^3+3x^2+8x+12)/(3x-4)$ ?

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Answer 1 · 2017-12-12T22:19:15

$4 x^{3} + x^{2} - 5$ $4x^3 +x^2-5$

So to begin, lets break down this problem to its lowest form

$\frac{4 \cdot x^{3} + 3 \cdot x^{2} + 8 \cdot x + 12}{3 \cdot x - 4} =$ $(4*x^3 +3*x^2+8*x+12)/(3*x-4)=$

We can eliminate any like terms, see any? How about

$(4*x^3 +cancel(3)*x^2+8*cancel(x)+12)/(cancel(3)*cancel(x)-4)=$

Now we can get rid of $-4$ by dividing it with the top

$(4*x^3 +x^2+cancel(20)^color(red)(-5))/cancel(-4)=$

So, we are left with $4x^3 +x^2-5$ and that is its simplest form! There you have it! Hope this helped!
~Chandler Dowd

How do you divide (4x^3+3x^2+8x+12)/(3x-4) 4x3+3x2+8x+123x−4(4x^3+3x^2+8x+12)/(3x-4) ?