# Thread: Dividing a polynomial by a binomial

1. ## Dividing a polynomial by a binomial

I have the process down pretty well but what I'm struggling with is when to change signs and exactly why.

For example: x^6 -13x^3 +42 divided by x^3 -7

I came up with X^3 +6 but in fact the answer is x^3 -6.

I have no problem writing out the entire problem if necessary but I don't want to make anyone do any laborious work in order to help out if it can be avoided.

Is there a straightforward answer that can be given as to when and why you change the signs?

2. Note that $x^6 - 13x^3 + 42$ is a quadratic in $x^3$, so make the substitution $X = x^3$ and it becomes

$X^2 - 13X + 42 = (X - 6)(X - 7) = (x^3 - 6)(x^3 - 7)$.

Therefore $\frac{x^6 - 13x^3 + 42}{x^3 - 7} = \frac{(x^3 - 6)(x^3 - 7)}{x^3 - 7} = x^3 - 6$.

3. Prove It's solution is the best. Although if you can't recognise this then you can use polynonimial long division with a little manipulation.

make $x^6 -13x^3 +42$ into $x^6 +0x^5+0x^4-13x^3 +0x^2+0x+42$

and $x^3-7$ into $x^3 +0x^2+0x-7$

then divide.

4. Thanks but we haven't been introduced to the word "quadratic" yet. We're supposed to be doing this problem in a long division format, which I get. I also believe that I get that the method you're showing me is more efficient but I have a test tomorrow and I have to go with what I know. Obviously I'm not exactly breaking boundaries in the world of mathematics so I need a really basic answer as to when and why I change the signs from negative to positive and vice versa to get the needed answer. I do not mean to sound unappreciative; quite the contrary. Thank you.

5. Pickslides, that is really great! My only question though is where does 13x^3 come into play? Or does it not? Do you just ignore it or am I [probably] missing something?

6. Originally Posted by Ingersoll
Thanks but we haven't been introduced to the word "quadratic" yet. We're supposed to be doing this problem in a long division format, which I get. I also believe that I get that the method you're showing me is more efficient but I have a test tomorrow and I have to go with what I know. Obviously I'm not exactly breaking boundaries in the world of mathematics so I need a really basic answer as to when and why I change the signs from negative to positive and vice versa to get the needed answer. I do not mean to sound unappreciative; quite the contrary. Thank you.
Sorry, but if you are doing long division of high order polynomials before you have been doing quadratics, then your school has a lot to answer for.

Quadratics are polynomials of degree 2, so they should be teaching you how to visualise, manipulate and transform these polynomials before moving onto anything higher.

7. Originally Posted by Ingersoll
Pickslides, that is really great! My only question though is where does 13x^3 come into play? Or does it not? Do you just ignore it or am I [probably] missing something?
From here?

Originally Posted by Ingersoll
I have the process down pretty well but what I'm struggling with is when to change signs and exactly why.

For example: x^6 -13x^3 +42 divided by x^3 -7

8. Okay, so you don't know about "quadratics" and are supposed to be doing "long division". Do you know how to do that? As pickslides said, you really need to "fill in" the missing terms:
$x^6 -13x^3 +42= x^6+ 0x^5+ 0x^4- 13x^3+ 0x^2+ 0x+ 42$
Now $x^3$ divides into $x^6$ " $x^3$ times so multiply each term of $x^3+ 0x^2+ 0x- 7$ by $x^3$ and subtract: $(x6+ 0x^5+ 0x^4- 13x^3+ 0x^2+ 0x+ 42)- (x^6+ 0x^5+ 0x^4- 7x^3)$. The $x^6$, $x^5$, and $x^4$ terms cancel leaving $(-13+ 7)x^3+ 0x^2+ 0x+ 42= -6x^3+ 0x^2+ 0x+ 42$. $x^3$ now divides into $-6x^3$ -6 times. -6 times $x^3+ 0x^2+ 0x- 7$ subtracted from $-6x^3+ 0x^2+ 0x+ 42$ is $-6x^3+ 0x^2+ 0x+ 42- (-6x^3+ 0x^2+ 0x+ 42)= 0$. That is, $x^3- 7$ divides into $x^6- 13x^3+ 42$ exactly $x^3- 6$ times.

9. Polynomial long division step1:

$\begin{array}{cclclcl}
& &x^3 \\ \cline{2-7}
x^3-7 & \vline & x^6&-&13x^3&+&42\\
& & x^6&-&7x^3\\ \cline{3-5}
& & & -&6x^3&+&42\\
\end{array}$

Step2:

$\begin{array}{cclclcl}
& &x^3& -&6\\ \cline{2-7}
x^3-7 & \vline & x^6&-&13x^3&+&42\\
& & x^6&-&7x^3\\ \cline{3-5}
& & & -&6x^3&+&42\\
& & & -&6x^3&+&42\\ \cline{4-7}
& & & && &0\\
\end{array}$

CB