Use synthetic division to divide the following problems. (x2 + 13x + 40) ÷ (x + 8)
Please help!!!!!
it is almost impossible to write division here start with the coefficients
1 13 40
then put a \(-8\) on the side 1 13 40 -8 ________________
bring down the 1 1 13 40 -8 _______________ 1
\(-8\times 1=-8\) 1 13 40 -8 -8 _______________ 1
add 1 13 40 -8 -8 _______________ 1 5
-8*5=-40 1 13 40 -8 -8 -40 _______________ 1 5
add 1 13 40 -8 -8 -40 _______________ 1 5 0
so there is no remainder and \[x^2+8x+40=(x+8)(x+5)\] which you probably could have figured out in ten seconds without division
oops typo there i meant \[x^2+13x+40=(x+8)(x+5)\]
thank you! would you be able to help me with one more? just to get me started because I think I got most of it. @satellite73
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Let f(x) = 4x3 + 7x2 – 13x – 3 and g(x) = x + 3. Find f of x over g of x
\[\frac{f(x)}{g(x)}=\frac{4x^3+7x^2-13x-3}{x+3}\] is a start now you can divide as before if you need to
this time you will get a quotient and a remainder, since \(x+3\) does not divide evenly
I totally agree with one of satelline's posts, why synthetic division when you can just factor ?
wanna use synthetic division for this one?
yes, please im supposed to for my assignment.
same as before list the coefficients 4 7 -13 3 -3 ____________________
bring down the 4 4 7 -13 3 -3 ____________________ 4
muliply -3*4=-12 4 7 -13 3 -3 -12 ____________________ 4
add 4 7 -13 3 -3 -12 ____________________ 4 -5
-3*-5=20 4 7 -13 3 -3 -12 20 ____________________ 4 -5
damn arithmetic mistake!
-3*-5=15!! 4 7 -13 3 -3 -12 15 ____________________ 4 -5 2
i see i made another mistake, the last number is \(-3\) not \(3\) so final answer is \[4x^2-5x-7-\frac{9}{x+3}\]
wait you lost me. where did the 20 come from?
that was a total mistake that is why i changed it to 15
oh okay sorry
there, just got rid of the mistake one and left the correct one up
okay! again thank you so much!!
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