Help with polynomials!! (1 Viewer)

[littlesmallworld]

given that (x-1) and (x+2) are two factors of 6x^4 +x^3 -17x^2 find a and b and the other two factors.
Help!! I already found a and b but i dont know how to find the other two factors

 

[rumbleroar]

Where is a and b in your equation?

 

[Mr Hippopotamus]

what?

 

[HeroicPandas]

Method 1: RELATIONSHIP BETWEEN ROOTS AND COEFFICIENTS
You know 2 roots of the polynomial, x = 1 and x = -2
Let the other 2 roots be A and B
Apply sum and product of roots, then solve simultaneously A and B

Method 2: LONG DIVISION
Long divide polynomial by (x-1)(x+2) = x² + x - 2
Write the polynomial again like this: 6x^4 +x^3 -17x^2 = (x^2 + x - 2 )(whatever u get from long division
Let polynomial equal to 0 to solve it
Then solve normally

 

[littlesmallworld]

Ummm.. My a is -1 and b is 16. The answers of the textbook say that the other two roots are (2x-1)(3x-2)

 

[MrBeefJerky]

Where is a and b? Are you sure they are factors? Subbing P (1) and P (-2) won't give you 0

 

[littlesmallworld]

Ohh right. Sorry. The origini equation was 6x^4 + ax^3 -17x^2 +bx -4.

 

[Mr Hippopotamus]

I have an extension maths test tomorrow and i have no clue how to do that....

I feel so bloody screwed...

 

[braintic]

Apply sum and product of roots, then solve simultaneously A and B

No need to do formal solution of simultaneous equations.

Just form a quadratic with roots A and B using A+B and AB found from sum & product:
x^2 - (A+B)x + AB = 0.
Then solve.

 

[MrBeefJerky]

No need to do formal solution of simultaneous equations.

Just form a quadratic with roots A and B using A+B and AB found from sum & product:
x^2 - (A+B)x + AB = 0.
Then solve.

Isn't it faster to just solve A and B and put it into the form (x-A)(x-B) than making it a quadratic equation and then using the quadratic formula which you still need to put it into that form?

 

[Verify]

When x=1 then f(x)=0 so there's one equation with both a and b.
When x=-2 then f(x)=0 so there's the other equation.

Then simultaneous equations. I think this is the fastest way.

 

[themoonspretty]

f(1)=0
6(1)^4+a(1)^3-17(1)^2+bx-4=0
a+b-15=0

f(-2)=0
6(-2)^4+a(-2)^3-17(-2)^2+b(-2)-4=0
96-8a-68-2b-4=0
24-8a-2b=0
12-4a-b=0

Now do simultaneous equations
a+b-15=0......1
and
12-4a-b=0.....2

From 2 b=12-4a sub into 1
a+12-4a-15=0
a=-1
Now sub a=-1 into b=12-4a
b=16.

Now
6x^4-x^3-17x^2+16x-4/(x-1)(x+2)

this equals 6x^2-7x+2 now factorise this
(2x-1)(3x-2)

 

[User5555555]

Wait, division of polynomials is in the 2 unit course? WHAT?!?

 

[Kurosaki]

Wait, division of polynomials is in the 2 unit course? WHAT?!?

It's not. given that this is in the 2013 exam discussion forum, it was probably put into the wrong section as to mathematics level as well, haha.

 

[Thank_You]

When x=1 then f(x)=0 so there's one equation with both a and b.
When x=-2 then f(x)=0 so there's the other equation.

Then simultaneous equations. I think this is the fastest way.

I think that's the best way