Look at 2x^3 + 5x^2 -11x + 4=0, and try to guess one solution, it is x=1, now divide 2x^3 + 5x^2 -11x + 4 with (x-1), you will get: 2 x^2+7 x-4
which is parabola and her solutions are x=-4, x=1/2
Hence: 2x^3 + 5x^2 -11x + 4=(x-1)(x+4)(2x-1)
How do you factor 2x^3 + 5x^2 -11x + 4? not quite sure where to start I can see that the last three terms make an ordinary quadratic and I know its in the right format with highest power first and thats about it.
The simplest approach is to start by looking for rational "zeroes" of the polynomial- rational number values of x that make . The "rational root theorem" says that if is a zero of the polynomial, then the numerator, m, must divide the constant term, 4, and the denominator, n, must divide the leading coefficient, 2. The only factors of 4 are , , and while the only factors of 2 are and .
That means that the only possible rational roots of the polynomial are , , and . We have to try each of those in the polynomial to see if one makes the polynomial equal to 0. Fortunately, we are successful immediately: [tex]2(1^3)+ 5(1^2)- 11(1)+ 4= 7- 11+ 4= 4- 4= 0.
That tells us that x- 1 is a factor. Now we can find another factor by long division, or its variant, synthetic division, or by "comparing coefficients": the other factor must be quadratic, of the form . Multiplying we get so we must have a= 2 and -c= 4. Then b-a= b- 2= 5 so b= 7. As a check c- b= -4- 7= -11.
Now we have [tex]2x^3+ 5x^2- 11x+ 4= (x- 1)(2x^2+ 7x- 4). We could continue to look at the possible rational roots , , and to find factors for but since it is quadratic we can use the quadratic formula: if then .
Those roots are not rational so if, by "factor", we mean "factor with integer coefficients", we cannot factor it any more: .
But in a more general sense,
If the coefficients (including 6) add up to 0, this means that 1 is one of the roots.
By factoring out (x - 1), then you're just left with solving a quadratic equation.
Question: wasn't this problem posted before on another thread?