multiplying binomial w/two trinomials

**greatsheelephant** · March 9th 2010, 05:06 AM

Sorry, I figured it out; I'd like to delete the thread can't seem to delete right now though.

**e^(i*pi)** · March 9th 2010, 05:10 AM

Originally Posted by greatsheelephant

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

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:
$-12y(4y^3+5y^2-12y)-2(4y^3+5y^2-12y)$
which gave me $-48y^4+76y^3-10y^2+24$
I then multiplied that by the other trinomial,
$(7y^2-9y+12)$
which gave me $-336y^6+964y^5-1330y^4+1002y^3+48y^2-216y+288$

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

Thanks

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

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

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

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