Question

Heres a simple x^3 one - i just forgot how to do them. f(x)=x^3 - 2x^2 - 13x - 10 as (x-a) (x-b) (x-c) [Simply the 3 factors for this equation]. (x+2) is one of them.

Answer 1

could someone answer an show as (x+2(x-b)(x-c) [usually (x-a)(x-b)(x-c) but using the FACTOR THEOREM]

Answer 2

to find the other factors your can either divide or think. you know it will be
\[x^3 - 2x^2 - 13x - 10(x+2)(\text{ something })\] and you can figure out the something by

Answer 3

i know thats the part im stuck at D:

Answer 4

by division, by synthetic division, or by pure reason. you know the leading term must be x^2 and the constant must be -5
then figure out what the middle term has to be

Answer 5

but you cannot use the factor theorem unless you know the zeros. if you know the zeros then you can use it to factor

Answer 6

\[x^3 - 2x^2 - 13x - 10=(x+2)(x^2 + \text{ something } -5)\] we know the -5 because the constant is -10, and we know the x^2 because the leading term is x^3
now we can reason out what "something" has to be because -5 times x is -5x but you have -13x so you need -8 x, which comes by multiplying 2 by -4x

Answer 7

I been using it by
\[x ^{3} - 2x ^{2} - 13x - 10 = (ax ^{2} + bx +c)(x+2)+r\]
\[x ^{3}+2x^{2}-13x-10=ax^{3} + 2ax^{2}+ bx^{2} +2bx +cx +2c + r\]
(Comparing x^3)
1=a
(Comparing x^2)
2=2a+2b
2=2+0
b=0
(Comparing x)
-13=2b+c
-13=0+c
c=-13
(Comparing 0)
-10=2c+r
-10=-26+r
r=16

so therefore
\[x ^{3} - 2x ^{2} - 13x - 10 = (x ^{2} + 0x -13)(x+2)+16\]
but from here i cant work it out >:/

Answer 8

oh sorry for Comparing x^2
2=2a+b