Synthetic Substitution

May 2010
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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)
 

Grandad

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Dec 2008
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Hello thisishard

Welcome to Math Help Forum!
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.SyntheticSubstitution.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 \(\displaystyle f(7) = 5\), and in the second \(\displaystyle f(2) = 5\). So work out what each of these expressions is in terms of \(\displaystyle k\), and then solve each equation to find \(\displaystyle k\).

Grandad
 

Grandad

MHF Hall of Honor
Dec 2008
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South Coast of England
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: \(\displaystyle f(-2) = 253\).
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 \(\displaystyle 2x^2+19x+24\), which factorises further into \(\displaystyle (2x+3)(x+8)\).
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 \(\displaystyle x = 7,\; 2x^2-8x+k = 154+k\). So, using the Remainder Theorem, if the remainder is \(\displaystyle 5\):
\(\displaystyle 154+k=5\)

\(\displaystyle \Rightarrow k = -149\)

(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.

Grandad
 

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May 2010
8
0
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
 

Grandad

MHF Hall of Honor
Dec 2008
2,570
1,416
South Coast of England
Hello 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 \(\displaystyle x^3+6x^2-x-30\) is divided by \(\displaystyle (x+5)\) the quotient is \(\displaystyle x^2+x-6\) (that's the meaning of the \(\displaystyle 1\; 1\; {-6}\)) and the remainder is \(\displaystyle 0\).

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

Grandad
 
May 2010
8
0
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
 

Grandad

MHF Hall of Honor
Dec 2008
2,570
1,416
South Coast of England
Hello thisishard

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

Grandad