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Factor theorum and remainder theorum

sunny1982

Hey folks I was wondering if someone can check these revision questions and see if they are correct please. thanks

1. Use the factor theorem to factorise completely f(x)= x³+x²-4x-4.
Hence solve the equation x³+x²-4x-4=0

x³+x²-4x-4=0
x² (x + 1) - 4(x + 1) = 0
(x + 2)(x + 1)(x - 2) = 0
x = -2 , -1 , 2

Remainder theorum questions
2. Determine the remainder when x³-6x²+x-5 is divided by:
a. X+2

f(x) = x³ - 6x² + x - 5
f(-2) = (-2)³ - 6(-2)² + (-2) - 5
-8 - 24 - 2 - 5 = -39

b. X-3

f(x)= x³ - 6x² + x-5
f(3)³ - 6(3)² + (3) - 5
27-54+3-5= -29

Is my remainder theorum correct now?

As for factor theorum I am lost.

(edited 11 years ago)

Reply 1

ghostwalker

Original post by sunny1982

Where'e the use of the factor theorem?

Remainder theorum questions
2. Determine the remainder when x³-6x²+x-5 is divided by:
a. X+2

f(x) = x³ - 6x² + x - 5
(x+2)= f(-2) = (-2)³ - 6(-2)² + (-2) - 5
-8 - 24 - 2 - 5 = -39

b. X-3

f(x)= x³ - 6x² + x-5
(x-3)=f(3)³ - 6(3)² + (3) - 5
27-72+3-5= -43

6 times 9 = ?

Reply 2

sunny1982

Original post by ghostwalker

Where'e the use of the factor theorem?

6 times 9 = 54 ?

is it meant to 54 instead of 72?

What do you mean I haven't used the factor theorum?

Reply 3

ghostwalker

Original post by sunny1982

is it meant to 54 instead of 72?

Yep.

What do you mean I haven't used the factor theorum?

Well you've clearly factorised the expression, but you've not used the factor theorem to do it. You jumped straight to the second part of the question and factorised f(x)=0 - which admittedly is easier to do in this situation.

Factor theorem says if f(a)=0 then f(x)=(x-a)g(x)

Reply 4

sunny1982

Original post by ghostwalker

Yep.

Well you've clearly factorised the expression, but you've not used the factor theorem to do it. You jumped straight to the second part of the question and factorised f(x)=0 - which admittedly is easier to do in this situation.

Factor theorem says if f(a)=0 then f(x)=(x-a)g(x)

So is my remainder theorem questions correct apart from the 54 obviously?lol

So which bit of the equation have I missed out on the factor theorum?

Reply 5

Joshmeid

Firstly find some value of 'a' for which f(a) = 0. Then take a factor of (x - a) from f(x), leaving g(x), then factorise the quadratic g(x).

Reply 6

ghostwalker

Original post by sunny1982

So is my remainder theorem questions correct apart from the 54 obviously?lol

Essentially.

(x+2)= f(-2) = (-2)³ - 6(-2)² + (-2) - 5

The bit in bold however doesn't make sense.

You could say "remainder on division by (x+2) is f(-2) = ...", of just "remainder is f(-2) = ..." since it's clear from the context what you're dividing by.

But (x+2) does not equal f(-2)

So which bit of the equation have I missed out on the factor theorum?

The start.

Initially you had f(x) =x³+x²-4x-4.

Then checking you find f(-1) = 0.

Hence f(x)=(x+1)g(x) and work out what g is.

And repeat for g(x).

It's not the easiest way to do it. And your method would probably be the best way to go about it if they hadn't said "Use the factor theorem".

This seems more an exercise in getting used to using the factor theorem, rather than just solving a polynomial.

Reply 7

sunny1982

Original post by Joshmeid

Firstly find some value of 'a' for which f(a) = 0. Then take a factor of (x - a) from f(x), leaving g(x), then factorise the quadratic g(x).

How?
I haven't got this in my notes :frown:

Reply 8

This Excellency

The factor theorem states if for some quadratric f(x), f(a)=0

Then (x-a) is a factor of f(x)

For your quadratic, show that f(1)=0

Reply 9

sunny1982

Remainder theorum questions
2. Determine the remainder when x³-6x²+x-5 is divided by:
a. X+2

f(x) = x³ - 6x² + x - 5
f(-2) = (-2)³ - 6(-2)² + (-2) - 5
-8 - 24 - 2 - 5 = -39

b. X-3

f(x)= x³ - 6x² + x-5
f(3)³ - 6(3)² + (3) - 5
27-54+3-5= -29

Is my remainder theorum correct now?

As for factor theorum I am lost.

1. Use the factor theorem to factorise completely f(x)= x³+x²-4x-4.
Hence solve the equation x³+x²-4x-4=0

x³+x²-4x-4=0
x² (x + 1) - 4(x + 1) = 0
(x + 2)(x + 1)(x - 2) = 0
x = -2 , -1 , 2

How many expressions am I missing and inbetween which line?

Reply 10

ghostwalker

Original post by sunny1982

Corrected the typo. Yes, that's fine.

As for factor theorum I am lost.

1. Use the factor theorem to factorise completely f(x)= x³+x²-4x-4.
Hence solve the equation x³+x²-4x-4=0

x³+x²-4x-4=0
x² (x + 1) - 4(x + 1) = 0
(x + 2)(x + 1)(x - 2) = 0
x = -2 , -1 , 2

How many expressions am I missing and inbetween which line?

It's the methodology.

By the factor theorem you show f(x)=(x + 2)(x + 1)(x - 2)

I'd suggest checking your textbook or Googling if you don't have any notes on the factor theorem.

THEN (this is the hence part of the question)

If f(x) = 0,
then (x + 2)(x + 1)(x - 2)=0

And so x = -2, or -1 or 2.

Reply 11

sunny1982

f(x) = x³ + x² - 4x - 4

f(x) = (x³ + x²) - (4x + 4)

f(x) = x²(x + 1) - 4(x + 1)

f(x) = (x + 1)(x² - 4)

f(x) = (x + 1)(x² - 2²)

f(x) = (x + 1)(x + 2)(x - 2)

x³ + x² - 4x - 4 = 0

(x + 1)(x + 2)(x - 2) = 0

I went a away and I've tried to factorise this completely using Factor theorum so I was wondering if this is correct or is there still some work need doing to it got an exam coming up :frown:

Reply 12

ghostwalker

Original post by sunny1982

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