Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
8741
3
avatar

Write the function in the form f(x) = (x − k)q(x) + r for the given value of k. f(x) = x3 + 5x2 − 3x − 22, k = sqrt(3)

 

 

 

Demonstrate that 
f(k) = r.

 Jul 28, 2016
edited by Guest  Jul 28, 2016
edited by Guest  Jul 28, 2016
edited by Guest  Jul 28, 2016

Best Answer 

 #3
avatar+26396 
+5

Write the function in the form f(x) = (x − k)q(x) + r for the given value of k.

f(x) = x3 + 5x2 − 3x − 22,

k = 3

 

We need to divide:

 

x3+5x23x22(x3)=x2+(5+3)x+53+7x3|(x3)x3+5x23x22=(x3)[ x2+(5+3)x+53 ]+(x3)[ 7x3 ]x3+5x23x22=(x3)[ x2+(5+3)x+53 ]=q(x)7=r

 

laugh

 Jul 28, 2016
 #1
avatar+9675 
0

f(x)÷(x3)=f(x)÷((x23)÷(x+3))=f(x)÷(x23)×(x+3)

 

We need to get f(x) into the form (x3)q(x)+r so we need to find q(x) and r

 

First we need to divide x^3 + 5x^2 - 3x - 22 by x^2-3

Quotient = x                           Remainder = -22

 

So that f(x)x23=x22x23

 

Then multiply this to x+sqrt3

 

(x22x23)(x+3)

=x2+3x22(x+3)x23

=x2+3x22x3

 

There we have found q(x) and r

q(x) = x^2 + sqrt3 x      r = -22

 Jul 28, 2016
 #2
avatar+33654 
+5

I think you forgot a 5x^2 in your subtraction Max!

 

Here's an alternative approach:

 

.

Alan  Jul 28, 2016
 #3
avatar+26396 
+5
Best Answer

Write the function in the form f(x) = (x − k)q(x) + r for the given value of k.

f(x) = x3 + 5x2 − 3x − 22,

k = 3

 

We need to divide:

 

x3+5x23x22(x3)=x2+(5+3)x+53+7x3|(x3)x3+5x23x22=(x3)[ x2+(5+3)x+53 ]+(x3)[ 7x3 ]x3+5x23x22=(x3)[ x2+(5+3)x+53 ]=q(x)7=r

 

laugh

heureka Jul 28, 2016

1 Online Users