# multiplying binomial w/two trinomials

• Mar 9th 2010, 05:06 AM
greatsheelephant
multiplying binomial w/two trinomials
Sorry, I figured it out; I'd like to delete the thread can't seem to delete right now though.
• Mar 9th 2010, 05:10 AM
e^(i*pi)
Quote:

Originally Posted by greatsheelephant
Here is the expression I'm supposed to find the product of:
\$\displaystyle (4y^3+5y^2-12y)(-12-2)(7y^2-9y+12)\$

To multiply it, does it matter whether or not I multiple a binomial with one of the trinomials first, or should I multiply the two trinomials together first? I can't seem to get it right. First I tried multiplying on eof the trinomials with the binomial:
\$\displaystyle -12y(4y^3+5y^2-12y)-2(4y^3+5y^2-12y)\$
which gave me \$\displaystyle -48y^4+76y^3-10y^2+24\$
I then multiplied that by the other trinomial,
\$\displaystyle (7y^2-9y+12)\$
which gave me \$\displaystyle -336y^6+964y^5-1330y^4+1002y^3+48y^2-216y+288\$

the correct answer is supposed to be \$\displaystyle -336y^6-44y^5+974y^4-1854y^3+1392y^2+288y\$

Thanks :)

\$\displaystyle -12-2 = -14\$ so that's the middle bracket sorted. Also you can take a factor of y from the first set.

\$\displaystyle -14y(4y^2+5y-12)(7y^2-9y+12)\$

Since this gives a polynomial of order 5 whereas the book says order 6 I suspect you've made a typo in the original question. Probably the middle bracket