# Thread: Factor & Remainder Theorem

1. ## Factor & Remainder Theorem

I have prob getting other two factors.
Had solved that (x-1) is a factor of 2X^3+5X^2-11X+4.

2X^3+5X^2-11X+4
=2(1)^3+5(1)^2-11(1)+4
=0

2. Originally Posted by s2951
I have prob getting other two factors.
Had solved that (x-1) is a factor of 2X^3+5X^2-11X+4.

2X^3+5X^2-11X+4
=2(1)^3+5(1)^2-11(1)+4
=0
Use long division to find the answer to $\frac{2x^3+5x^2-11x+4}{x-1}$

This should give a quadratic from which you can find the other two solutions (remember that $x-1$ is also a factor)

3. I prefer to do this method, but got stuck.. whats next?

f(x)=(x-1)q(x)
2x^3+5x^2-11x+4=(x-1)(2x^2+x-4)
2x^3+5x^2-11x+4=x(2x^2+x-4)-1(2x^2+x-4)
??? = ???

4. Originally Posted by s2951
I prefer to do this method, but got stuck.. whats next?

f(x)=(x-1)q(x)
2x^3+5x^2-11x+4=(x-1)(2x^2+x-4)
2x^3+5x^2-11x+4=x(2x^2+x-4)-1(2x^2+x-4)
??? = ???
You appear to be assuming that the other factor is $2x^2+ x- 4$. Why?

If you insist upon that method (though ordinary division or synthetic division is simpler) you need to write $2x^3+ 5x^3- 11x+ 4= (x-1)(ax^2+ bx+ c)$. Do the multiplication and combine like powers and you will get three equations for a, b, and c. (Actually, you get 4 equations but they are not "independent" since x-1 is a factor of $2x^3+ 5x- 11x+ 4$.

5. Originally Posted by s2951
I prefer to do this method, but got stuck.. whats next?

f(x)=(x-1)q(x)
2x^3+5x^2-11x+4=(x-1)(2x^2+x-4)
2x^3+5x^2-11x+4=x(2x^2+x-4)-1(2x^2+x-4)
??? = ???
Let $q(x)=Ax^2+Bx+C$ (it's a method I prefer)

$(x-1)(Ax^2+Bx+C) = 2x^3+5x^2-11x+4$

Compare coefficients

• $x^3 \: \rightarrow \: 2 = A$
• $x^2 \: \rightarrow \: 5 = -A+B \: \: \therefore \: B = 7$
• $x^0 \: \rightarrow \: 4 = -C$

$(x-1)(2x^2+7x-4) = 2x^3+5x^2-11x+4$

$\Delta = 7^2-(4 \times 2 \times -4) = 81$. As this is a perfect square the quadratic has rational solutions and can be factorised.

$2x^2+7x-4 = (2x-1)(x+4)$

Therefore $2x^3+5x^2-11x+4 = (x-1)(x+4)(2x-1)$