how to factor four terms

**anthonye** · October 7th 2010, 03:21 AM

Hi all;
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.

Thanks.

**Also sprach Zarathustra** · October 7th 2010, 03:28 AM

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)

**HallsofIvy** · October 7th 2010, 03:46 AM

The simplest approach is to start by looking for rational "zeroes" of the polynomial- rational number values of x that make $2x^3+ 5x^2- 11x+ 4= 0$ . The "rational root theorem" says that if $\frac{m}{n}$ 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 $\pm 1$ , $\pm 2$ , and $\pm 4$ while the only factors of 2 are $\pm 1$ and $\pm 2$ .

That means that the only possible rational roots of the polynomial are $\pm 1$ , $pm 2$ , $\pm 4$ and $\pm \frac{1}{2}$ . 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 $ax^2+ bx+ c$ . Multiplying $(x- 1)(ax^2+ bx+ c)$ we get $ax^3+ bx^2+ cx- ax^2- bx- c= ax^3+ (b- a)x^2+ (c- b)x- c= 2x^3+ 5x^2- 11x+ 4$ 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 $\pm 1$ , $\pm 2$ , $\pm 4$ and $\pm \frac{1}{2}$ to find factors for $2x^2+ 7x- 4$ but since it is quadratic we can use the quadratic formula: if $2x^2+ 7x- 4= 0$ then $x= \frac{-7\pm\sqrt{7^2- 4(2)(-4)}}{2(2)}$ $= \frac{-7\pm\sqrt{40+ 32}}{4}= \frac{-7\pm\sqrt{72}}{4}=$ $\frac{-7\pm 6\sqrt{2}}{4}$ .

Those roots are not rational so if, by "factor", we mean "factor with integer coefficients", we cannot factor it any more: $2x^3+ 5x^2- 11x+ 4= (x- 1)(2x^2+ 7x- 4)$ .

But in a more general sense, $2x^3+ 5x^2- 11x+ 4= 16(x- 1)\left(x- \frac{7}{4}+\frac{3\sqrt{2}}{2}\right)\left(x-\frac{7}{4}-\frac{3\sqrt{2}}{2}\right)$

**Also sprach Zarathustra** · October 7th 2010, 03:55 AM

7^2=49

**wonderboy1953** · October 7th 2010, 08:27 AM

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?

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