Factorize 2x^3 + x^2 - 13x + 6?

Hi,

Since this equation could have roots of ±1,±2,±3, or ±6 as well as fractions of ±1/2 or ±3/2, try synthetic division by one of these numbers looking for a zero remainder.
. ___________
1)2 1 -13. . 6
.__2__3_-10_
. 2. 3 -10 . -4 <=with a remainder of -4, 1 is not a root

. ___________
2)2 1 -13 . 6
.__4_10_-6_
. 2. 5. -3 . 0 <=with a remainder of 0, so x = 2 is a root and x - 2 is a factor

2x³ + x² - 13x + 6 factors into (x - 2)(2x² + 5x - 3)

(2x² + 5x - 3) factors into (2x - 1)(x + 3)

So, 2x³ + x² - 13x + 6 factors into (x - 2)(2x - 1)(x + 3) <==ANSWERS

I hope that helps!! :-)

The remainder theorem means that when you substitute a number into the equation and it works out to zero that it is a factor. Hence the 0.5, 2, -3.

To do this, you have to use a technique called long division, in which you can substitute the numbers. so if (x+3) is a factor for example, x=-3 by transposition.

Alternatively you can ask your maths teacher how to use a cas calculator, which is super useful. But still, LEARN LONG DIVISION, ITS IMPORTANT

@Cara Cara No it isn't, I have used a cas and it works out fine.

To find the first factor it is basically just trial and error.. try 1, -1, 2, -2, etc.

When you get one that makes the whole equation = 0 (like 2) then you have found one factor, in this case, x = 2. Now you can do this for all three factors or you can then use synthetic or long division.]

Hope I helped! If you don't know how to do either of these, could you edit your question?

the last answer is wrongg.