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Can somebody help me with this question?

$$\frac{15p^3+16p^2+46}{3p+5}$$

For some reason I can't wrap my head around the process used to divide polynomials, I can do long division but every time somebody explains the long division of polynomials to me I can't understand it whatsoever.

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    $\begingroup$ I don't understand. You say that you can do long divisions, but if someone try to explain them to you, you don't undertand. Then, how can you do them? $\endgroup$ – ajotatxe Jul 22 '15 at 0:34
  • $\begingroup$ Normal long division. I can't comprehend for the life of me the process of dividing polynomials like the one I've posted, I don't know why. $\endgroup$ – user256027 Jul 22 '15 at 0:36
  • $\begingroup$ Can you do the first step? Look at the leading terms of numerator and denominator, and find a "trial quotient" by dividing them. $\endgroup$ – hardmath Jul 22 '15 at 0:36
  • $\begingroup$ @user256027 Are you wanting to know why polynomial long division works? $\endgroup$ – Christopher Jul 22 '15 at 0:41
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Here's another way of dividing them. I think it's easier to remember than long division (although it's essentially the same).

The first step: \begin{align} \frac{15p^3+16p^2+46}{3p+5}&=\frac{15p^3+25p^2-25p^2+16p^2+46}{3p+5} \\&=\frac{5p^2(3p+5)-25p^2+16p^2+46}{3p+5}\\&= \frac{5p^2(3p+5)}{3p+5}+\frac{-25p^2+16p^2+46}{3p+5}\\&=5p^2+\frac{-25p^2+16p^2+46}{3p+5}\\ &= 5p^2+\frac{-9p^2+46}{3p+5}\end{align} To sum up, you look at the term on the numerator with the largest exponent of $p$, here $25p^3$, and the term in the denominator with the largest exponent of $p$, here $3p$, and you add and subtract a number that will allow you to factor out the whole denominator.

So in the first step we saw that $15p^3=5p^2(3p)$, and then we used the $5p^2$ to decide what number to add and subtract: $5p^2(3p+5)=15p^3+25p^2$. We already have the $15p^2$, but we're missing the $25p^2$, so this is the number we'll add and subtract (if the denominator had more terms, we would add and subtract more terms).

You can do this with each term on the numerator until you get a term whose degree (the largest exponent of $p$ with nonzero coefficient) is smaller than that on the denominator. In this case, since the degree of $3p+5$ is $1$, your remainder will be a constant.

Try performing the other steps:

  • to factor out $-3p$ you will add and subtract $-15p$
  • to factor out $5$, you will add and subtract $25$
  • you'll be left with a remainder of 21
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$\color{red}{\rm Reds}$ are the terms we want to kill, $\color{blue}{\rm blues}$ are pieces of our answer.

Consider $\color{red}{15p^3}+16p^2 + 46$. If I multiply $3p+5$ by $\color{blue}{5p^2}$, we get $\color{red}{15p^3}+25p^2$.

Subtracting it, we're left with $\color{red}{-9p^2}+46$. If I multiply $3p+5$ by $\color{blue}{-3p}$, we get $\color{red}{-9p^2}-15p$.

Subtracting it, we're left with $\color{red}{15p}+46$. If I multiply $3p+5$ by $\color{blue}{5}$, we get $\color{red}{15p}+25$.

Subtracting it, we're left with $\color{green}{24}$. Since the degree of $24$ is less than the degree of $3p+5$, the process ends.

$$15p^3+16p^2 + 46 = (\color{blue}{5p^2-3p+5})(3p+5)+\color{green}{24}.$$

How we find our $\color{blue}{\rm blue}$ terms? Look at the leading terms. In each step: $$\frac{\color{red}{15p^3}}{3p} = \color{blue}{5p^2}, \quad \frac{\color{red}{-9p^2}}{3p} = \color{blue}{-3p}, \quad \frac{\color{red}{15p}}{3p} = \color{blue}{5}.$$

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Step one: Find $x$ such that the first term of your divisor ($3p$) times x equals the first term of the dividend $15p^3$ (it is $x=5p^2$).

Step two: multiply $3p+5$ by $x$ (here we get $15p^3-25p^2$)

step three: subtract the result from step 2 from your dividend.

Step four: Repeat steps 1-3 with your result from step 3 until the power of $p$ in your divisor (here it is 1) exceeds the power of $p$ in your result from step 3. (what is left is called the remainder)

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There are two ways to approach this.

The "long" way is to do polynomial long division. It's not that difficult if you just consider the lead coefficients when you're deciding the "how much into" part, but take the whole expression into account when subtracting. Rather than trying (and very likely failing) to typeset the whole thing in my rudimentary Latex, I'll just link you to this: http://www.purplemath.com/modules/polydiv2.htm

The "short" way, which is what I'll highlight here, is something called synthetic division, which works only when you have linear (degree one) factors in the divisor. In this case, this works fine, but you have to ensure the lead coefficient of the divisor is reduced to one by dividing throughout by $3$. Here's the link for synthetic division: http://www.purplemath.com/modules/synthdiv.htm

You might find synthetic division much easier because you're dealing only in numbers (the coefficients) and adding rather than subtracting. Keep in mind that if you divided both numerator and denominator by a constant to get a monic linear factor in the divisor (for synthetic division), your quotient will not be affected but your remainder will. So you must multiply by that constant to get the remainder for the original division problem.

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A way I like to look at polynomial long division is by realising that 3p+5 is a factor for all real values of p.

Thus, you can say that the quotient is equal to some polynomial ap^2+bp+c. I got the degree by subtracting the degree of the numerator from the degree of the denominator, which makes sense if you multiply the denominator over to the other side when you're done.

By plugging in values for p, I can obtain simultaneous equations in a,b and c, which can be solved. You can find the c term right away by plugging in zero for p. Also the value of a can be found by dividing the coefficients of the highest degree terms in numerator and denominator.

I personally found this technique straightforward enough, and doesn't require so much polynomial multiplication. The only problem is that it does assume that the denominator is a factor of the numerator, and doesn't really work if it's not.

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