Question

Factor polynomial. One factor is given.
F(x)=3x^3+33x^2-36
Factor: x+12

Answer 1

\[x^3+33x^2-36=(x+12)(\text{some quadratic})\]

Answer 2

you can find the quadratic by
a) long division (ick)
b) synthetic division (very easy) or
c) thinking what it has to be

Answer 3

How is it done like how to do it step by step

Answer 4

i cannot write synthetic division here, i have tried and it is always a mess
do you know how to do it?

Answer 5

I need to refresh my memory

Answer 6

we can think if you like

Answer 7

\[x^3+33x^2-36=(x+12)(ax^2+bx+c)\] it should be pretty obvious that the first term has to be \(x^2\) because you have \(x^3\) when you multiply on the right
it also should be obvious that the last term has to be \(+3\) because that is how you are going to get \(-36\) so \[x^3+33x^2-36=(x+12)(x^2+bx-3)\] we need \(b\) only

Answer 8

i meant the last term has to be \(-3\) sorry

Answer 9

so all you need is \(b\) and we can also find that by thinking

Answer 10

`3x^3+33x^2-36` is really `3x^3+33x^2+0x-36`
write out the coefficients in a horizontal line
then place -12 (the test zero; found by solving x+12 = 0) to the left of the list of coefficients