+0  
 
+3
922
8
avatar+63 

I forgot how to multiply polynomials and a problem that requires this skill showed up on my hw that it due tomorrow. 

 Feb 13, 2019
 #1
avatar+36915 
0

Bummer for you , Dude......  cheeky

 Feb 13, 2019
 #2
avatar+63 
+3

I know 

TheDeathlyHallows  Feb 13, 2019
 #3
avatar+36915 
0

If you have  a sample, someone can show you how it is done,.......

ElectricPavlov  Feb 13, 2019
 #4
avatar+63 
+3

Ok, the problem I need solved is (4x+17)*(3x+2)

TheDeathlyHallows  Feb 13, 2019
edited by TheDeathlyHallows  Feb 13, 2019
 #5
avatar+36915 
0

Multiply  4x  times each of the terms in the second factor...then multiply 17 times each of the terms in the second factor...

    then collect like terms:

 

4x*3x + 4x * 2  + 17 * (3x) + 17 * 2

12x^2  +  59x + 34                                  Just be careful when there is a negative sign involved!

ElectricPavlov  Feb 13, 2019
 #6
avatar+63 
+3

thanks so much! It really helped me 

TheDeathlyHallows  Feb 13, 2019
 #7
avatar+63 
+3

I thought I was done and apparently, I have a division problem. (x^2-5x+6)/((2x-1)+(x/2)) is the problem. 

TheDeathlyHallows  Feb 13, 2019
 #8
avatar+2439 
+2

Hey, DeathlyHallows!

 

It looks like you want to simplify \(\frac{x^2-5x+6}{\textcolor{red}{2x-1+\frac{x}{2}}}\) . Let's transform all the terms in the denominator such that they all have common denominators; this way, we will be able to add them together. 

 

\(\textcolor{red}{2x-1+\frac{x}{2}}\Rightarrow\textcolor{red}{\frac{4x}{2}-\frac{1}{2}+\frac{x}{2}}\)

 

Notice that I have not changed the value of the expression; I just created a common denominator. Now, combine like terms. 

 

\(\textcolor{red}{\frac{4x}{2}-\frac{1}{2}+\frac{x}{2}\\ \frac{5x}{2}-\frac{1}{2}\\ \frac{5x-1}{2}}\)

 

Therefore, \(\frac{x^2-5x+6}{2x-1+\frac{x}{2}}=\frac{x^2-5x+6}{\frac{5x-1}{2}}\) . Now, it is time to simplify. 

 

\(\frac{x^2-5x+6}{\frac{5x-1}{2}}*\frac{2}{2}\\ \frac{2(x^2-5x+6)}{5x-1}\\ \frac{2(x-2)(x-3)}{5x-1}\)

 

Even after I factored the numerator completely, I did not find any common factors present in both the numerator and the denominator, so this expression is already in simplest form. 

TheXSquaredFactor  Feb 16, 2019

2 Online Users

avatar
avatar