Thread: 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

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

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

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)
??? = ???

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 . Why?

If you insist upon that method (though ordinary division or synthetic division is simpler) you need to write . 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 .

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 (it's a method I prefer)



Compare coefficients

• 
• 
• 

. As this is a perfect square the quadratic has rational solutions and can be factorised.



Therefore