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Factor And Remainder Theorem

Questions

6. When the polynomial 3x^2+ax^2+bx-1 is divided by x-2 the remainder is 5. When it is divided by x+1 the remainder is -1.
Find the values of a and b.

7. Solve the equations:
a) 3x^3-2x^2-11x+10=0
b) 2x^3+5x^2-14x-8=0
c) 4x^3+12x^2-7x-30=0

I have been studying how to use the factor theorem and algebraic long division, I was set these questions as part of a 2 hour assignment and I haven't got a clue on what to do.

I don't want solutions, I want to learn how to correctly work these out.
I have spent around 30 mins trying methods on each question with no luck. Kinda why I'm asking for help
Reply 2
what have you done so far?

tell me what the remainder theorem means, in your own words.
Original post by Thahleel
Questions

6. When the polynomial 3x^2+ax^2+bx-1 is divided by x-2 the remainder is 5. When it is divided by x+1 the remainder is -1.
Find the values of a and b.

7. Solve the equations:
a) 3x^3-2x^2-11x+10=0
b) 2x^3+5x^2-14x-8=0
c) 4x^3+12x^2-7x-30=0

I have been studying how to use the factor theorem and algebraic long division, I was set these questions as part of a 2 hour assignment and I haven't got a clue on what to do.

I don't want solutions, I want to learn how to correctly work these out.

For question 6, you can subtract the equation by 5 in order to make x-2 become a factor of the equation you got and minus -1 in order to make x+1 become a factor of the equation you got. Then use the factor theorem to get a simultaneous equation.
Well what I've been told by the teacher is to use the factor theorem to work out the polynomials, no number seems to work.

I know very little on the remainder theorem, been told by my teacher to use the factor theorem to work out remainders instead of using full blown long division.

I factor f(2) into 3x^2+ax^2+bx-1=5 and I don't really know what to do next since theres 2 values I have to work out.

For the other 3 questions I've tried factoring in possible numbers so for 7a) I've used f(5), f(-5), f(10), f(-10) with them all returning remainders.

I'd like to know where I've gone wrong.
Original post by bird in water
For question 6, you can subtract the equation by 5 in order to make x-2 become a factor of the equation you got and minus -1 in order to make x+1 become a factor of the equation you got. Then use the factor theorem to get a simultaneous equation.


That's where I got to. F(2) = (a*4)+(b*2)=-18
And a-b=3

Where do I go from that?
Reply 6
Hint for 7a

Try f(1)
Original post by steve2005
Hint for 7a

Try f(1)


So how does that help me find the other 2?
Original post by Thahleel
That's where I got to. F(2) = (a*4)+(b*2)=-18
And a-b=3

Where do I go from that?

you can get a=3+b from a-b=3
then substitute a in the first equation
Reply 9
Original post by Thahleel
So how does that help me find the other 2?


Since f(1) = zero you know that x-1 is a factor, this leaves you with a quadratic.
Original post by bird in water
you can get a=3+b from a-b=3
then substitute a in the first equation


Oh I see. This is touching on that stuff I was taught about simultaneous equations for GCSE a few years ago
Original post by steve2005
Since f(1) = zero you know that x-1 is a factor, this leaves you with a quadratic.


Then I've got to use Long Division to find the Quadratic?
Original post by Thahleel
So how does that help me find the other 2?

you can use the algebraic long division to get a quadratic equation and then just simply solve it!
Thanks for the help guys, I really appreciate it.
For question 6 do the same again using -1 instead of 2, then use your two answers and solve as a simultaneous equation :smile:
Reply 15
Original post by floydmarshy
For question 6 do the same again using -1 instead of 2, then use your two answers and solve as a simultaneous equation :smile:


I think you've just resurrected a 2 year old thread! :smile:

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