1. ## Synthetic Substitution

I missed alot of school so i dont know how to do this

use synthetic substitution to find f(-2). show all work

f(x) = -9x^5 + 5x^3 - 2x + 1

f(x) = 3x^3 - 4x + 6

given a polynomial and one of its factors, find the remaining factors of the polynomial. some factors may not be binomials.

2x^3 + 15x^2 - 14x -48 ; x - 2

x^3 + 6x^2 - x - 30 ; x + 5

Find values of k so that each remainder is 5

(2x^2 - 8x + k) / (x - 7)

(x^4 + kx^3 - 7x^2 + 8x + 25) / (x - 2)

2. Hello thisishard

Welcome to Math Help Forum!
Originally Posted by thisishard
I missed alot of school so i dont know how to do this

use synthetic substitution to find f(-2). show all work

f(x) = -9x^5 + 5x^3 - 2x + 1

f(x) = 3x^3 - 4x + 6
Have you tried Googling 'synthetic substitution'? I found this very clear explanation very easily: http://hanlonmath.com/pdfFiles/29.Sy...bstitution.pdf

given a polynomial and one of its factors, find the remaining factors of the polynomial. some factors may not be binomials.

2x^3 + 15x^2 - 14x -48 ; x - 2

x^3 + 6x^2 - x - 30 ; x + 5
And I Googled 'synthetic division' and came up with: Synthetic Division
Find values of k so that each remainder is 5

(2x^2 - 8x + k) / (x - 7)

(x^4 + kx^3 - 7x^2 + 8x + 25) / (x - 2)
Use the remainder theorem. In the first of these $f(7) = 5$, and in the second $f(2) = 5$. So work out what each of these expressions is in terms of $k$, and then solve each equation to find $k$.

3. I still don't understand!!!!

4. ## Synthetic Substitution and Division

Hello again thisishard!

I can't explain synthetic substitution and division any more clearly than in the articles that I directed you to.

It is difficult to set the working out using LaTeX so I've done it by hand, and attached image files.

use synthetic substitution to find f(-2). show all work

f(x) = -9x^5 + 5x^3 - 2x + 1
The working for this is shown in the attachment I've called SyntheticSubstitution.jpg. The answer: $f(-2) = 253$.
f(x) = 3x^3 - 4x + 6
I hope you can do this one now.
given a polynomial and one of its factors, find the remaining factors of the polynomial. some factors may not be binomials.

2x^3 + 15x^2 - 14x -48 ; x - 2
I've done the division using ordinary long division, and also synthetic long division. The working is shown in Division.jpg. The other factor, therefore, is $2x^2+19x+24$, which factorises further into $(2x+3)(x+8)$.
x^3 + 6x^2 - x - 30 ; x + 5
I'll leave this one for you to try.
Find values of k so that each remainder is 5

(2x^2 - 8x + k) / (x - 7)
When $x = 7,\; 2x^2-8x+k = 154+k$. So, using the Remainder Theorem, if the remainder is $5$:
$154+k=5$

$\Rightarrow k = -149$
(x^4 + kx^3 - 7x^2 + 8x + 25) / (x - 2)
Again, I'll leave you to have a go at this.

If you want us to confirm that you have the remaining answers right, get back to us.

5. given a polynomial and one of its factors, find the remaining factors of the polynomial. some factors may not be binomials.

x^3 + 6x^2 - x - 30 ; x + 5

this would be
-5_| 1 6 -1 -30
-5 -5 30
------------------
1 1 -6 0

would that be the whole answer because it doesn't seem like it

6. Hello thisishard
Originally Posted by thisishard
given a polynomial and one of its factors, find the remaining factors of the polynomial. some factors may not be binomials.

x^3 + 6x^2 - x - 30 ; x + 5

this would be
-5_| 1 6 -1 -30
-5 -5 30
------------------
1 1 -6 0

would that be the whole answer because it doesn't seem like it
Your working shows that when $x^3+6x^2-x-30$ is divided by $(x+5)$ the quotient is $x^2+x-6$ (that's the meaning of the $1\; 1\; {-6}$) and the remainder is $0$.

So you have shown that
$x^3+6x^2-x-30=(x+5)(x^2+x-6)$
Now you can factorise the second term further into $(x+3)(x-2)$. So $(x+3)$ and $(x-2)$ are the remaining factors of the original polynomial.

7. Huh??? second term?

8. I learn best if i see some one do a problem completly so will you please finish it its the only part im stuck on

I already did the rest thanks to you

9. Hello thisishard

I don't really see where the problem is. We have shown that
$x^3+6x^2-x-30=(x+5)(x^2+x-6)$
The second term on the right-hand-side - that is, $(x^2 +x-6)$ - can be factorised:
$x^2+x-6 = (x+3)(x-2)$
So $(x+3)$ and $(x-2)$ are the remaining factors of $x^3+6x^2-x-30$.

10. where did the right side come from?

11. Hello thisishard
Originally Posted by thisishard
where did the right side come from?
The synthetic division you did, that gave the result 1 1 -6 0, means that when $x^3+6x^2-x-30$ is divided by $(x+5)$ the answer (the quotient) is $\color{red}(1)\color{black}x^2+\color{red}(1)\colo r{black}x\color{red}-6$ and the remainder is $\color{red}0$. In other words:
$\frac{x^3+6x^2-x-30}{x+5} =x^2+x-6$
So:
$x^3+6x^2-x-30=(x+5)(x^2+x-6)$
OK?