+0

# how do you do a(2a+3)-a(a+1) I got 3a + a squared + 1 Is that right

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how do you do a(2a+3)-a(a+1) I got 3a + a squared + 1 Is that right

May 7, 2014

#2
+121064
+5

how do you do a(2a+3)-a(a+1) I got 3a + a squared + 1 Is that right

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You're close......let's look at it again.

Distributing the "a" across 2a + 3 in the first expression gives us 2a2 + 3a

And distributing the "-a" across a + 1 in the second expression gives us -a2 - a

So, combining like terms, we have....... a2 + 2a

(I think you might have forgotten to distribute the "-a" across the +1 in the second part.)

May 7, 2014

#1
+1006
0

First, distribute the 'a's.

$$\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{a}}\right){\mathtt{\,-\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{a}}\right)$$

What will happen now is that, in order to factor the polynomials, you must set up a 3x3 grid and combine (in a similar fashion to Punnett Squares).

(I have to go now, can somebody finish this one up?)

May 7, 2014
#2
+121064
+5

how do you do a(2a+3)-a(a+1) I got 3a + a squared + 1 Is that right

-----------------------------------------------------------------------------------------------------------------------------

You're close......let's look at it again.

Distributing the "a" across 2a + 3 in the first expression gives us 2a2 + 3a

And distributing the "-a" across a + 1 in the second expression gives us -a2 - a

So, combining like terms, we have....... a2 + 2a

(I think you might have forgotten to distribute the "-a" across the +1 in the second part.)

CPhill May 7, 2014