SOLUTION: I'm not sure if this is the right section, but I will still ask. 1. Use synthetic division to find f(-3) for f(x)2x^3-6x^2-5x+7. 2. One factor of f(x)=x^3+x^2-22x-40 is x+4. F

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I'm not sure if this is the right section, but I will still ask. 1. Use synthetic division to find f(-3) for f(x)2x^3-6x^2-5x+7. 2. One factor of f(x)=x^3+x^2-22x-40 is x+4. F      Log On


   

Question 166180: I'm not sure if this is the right section, but I will still ask.
1. Use synthetic division to find f(-3) for f(x)2x^3-6x^2-5x+7.
2. One factor of f(x)=x^3+x^2-22x-40 is x+4. Find the other factors.
I must show my work and have no textbook to explain. Will you PLEASE HELP!!!!
Thanks
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1. You don't need synthetic division to find f(-3). Just substitute.
f%28x%29=2x%5E3-6x%5E2-5x%2B7
f%28-3%29=2%28-3%29%5E3-6%28-3%29%5E2-5%28-3%29%2B7
f%28-3%29=2%28-27%29-6%289%29%2B15%2B7
f%28-3%29=-54-54%2B15%2B7
f%28-3%29=-86
.
.
.
2. Here you need the synthetic division.
Showing synthetic division is a little difficult.
The first part will show the factor (left hand column), then the factor times the divisor,
then the next line will show the subtraction.
Repeat.
Hopefully it makes sense.
.
.
.
..........._______________
(x+4)|x^3+x^2-22x-40
x^2:...x^3+4x^2
..................-3x^2-22x-40
-3x:..........-3x^2-12x
.............................-10x-40
-10:......................-10x-40
x%5E3%2Bx%5E2-22x-40=%28x%2B4%29%2A%28x%5E2-3x-10%29
Now you can factor the quadratic,
x%5E2-3x-10=%28x-5%29%28x%2B2%29
and substitute back,
x%5E3%2Bx%5E2-22x-40=%28x%2B4%29%2A%28x-5%29%2A%28x%2B2%29