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# Quotient and remainder

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(6a^3+11a^2-4a-9)÷(3a-2)

Feb 2, 2022

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I brute forced the problem. $${{6a^3}\over {3a } } = 2a^2$$... and so on, until I found the quotient, which is the trinomial: $$\color{brown}\boxed{2a^2+5a+2}$$.

When you multiply this by $$3a-2$$, you get: $${6a^3+11a^2-4a-4}$$. The remainder to this problem is $$\color{brown}\boxed{-5}$$

(I'll explain it more in-depth in 45 min once I get home.)

Feb 2, 2022
edited by BuilderBoi  Feb 2, 2022
#2
+1334
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$$(6a^3+11a^2-4a-9)\over(3a-2)$$

At the end, it will be $$(3a-2)(xa^2+ya+z) + r$$

In this equation, $$x$$$$y$$, and $$z$$ are coeficents, and $$r$$ is the remainder.

We know that x must equal 2, because $$2a^2(3a-2)=6a^3-4a^2$$

We know that y equals 5, because (note we already have a -4 so it will add to 11): $$5a(3a-2) = 15a^2-10a$$

We know that z must equal 2, because(Note: we already have a -10, so it will add to -4): $$2(3a-2) = 6a-4$$.

When we add everything, we have: $$6a^3+11a^2-4a-4$$. This means that the remainder is $$\color{brown}\boxed{-5}$$, and the quotient is $$\color{brown}\boxed{2a^2-5a+2}$$

Feb 2, 2022