How do i factor something with that has a cubic power in it?

Question

how do you factor 2x^3+x^2+1

explanation? because im quite confused.

Answer 1

The main way is to guess a root r. Then x - r will be a factor. Divide that in, and you're left with a quadratic,which may or may not factor further.

A hugely important point is that r, if it exists, will divide the constant term of the cubic polynomial. That greatly restricts the guessing stage. In this problem, r has to divide 1, so there are only two guesses --1 and -1.

To see WHY r has to divide the constant term, just formally multiply out (x-r)(x^2+ax+b), and look at the constant term. It will be -br, which of course is divisible by r.

In this case, the polynomial has a root at -1, where it equals -2+1+1=0. So it can be divided by x+1. It turns out to factor into (x+1)(x^2+x-1). And since x^2+x-1 happens to be irreducible, that's indeed a complete factorization.

Answer 2

Find one root, factor it out and then factor the resulting quadratic (if possible). In this case:

f(x) = 2x³ + x² + 1
f(-1) = -2 + 1 + 1 = 0
x+1 is a factor

2x³ + x² + 1
= (x+1)(2x²-x+1)

(2x²-x+1) cannot be factored any further in the real domain

2x³ + x² + 1
= (x+1)(2x²-x+1)

Answer 3

2x^3 +x^2 + 1

you write 2x^3 as x^3 + x^3 and take out x^2 as a factor from x^3 and x^2, the remaiming terms are x^3+ 1, which can be factored using (a^3+b^3= (a+b)(a^2-ab+b^2))

2x^3 + x^2 + 1 = x^3 + x^3 + x^2 + 1

=>x^3 + x^2 + x^3+1

=>x^2(x+1)+ (x+1)(x^2 - x + 1)

=>(x+1)(x^2+x^2-x+1)

=>(x+1)(2x^2- x + 1)

Another method is using factor theorem. If x = a root
then f(a) = 0.

f(x) = 2x^3 + x^2 + 1

f(-1) = 2(-1)^3 +(-1)^2 + 1

f(-1) = -2 + 1 + 1 = 0

so x=-1 is a root and x+1 is a factor.Now using long division you divide f(x) by (x+1).

x+1)2x^3+x^2+0x+1(2x^2
___2x^3+2x^2
______________
_______-x^2+0x(-x
_______-x^2-x
______________
__________x+1(1
__________x+1
_____________
___________0

so the quotient is 2x^2-x+1

2x^3+x^2+ 1 = (x+1)(2x^2-x+1)

Answer 4

try different values of x +/- a until something gives you a zero then practice factor long division to get you a quadratic equation which you can do using the standard formula

y = 2x^3 + x^2 + 0x + 1

Y(0) = 1 so discard x = 0

y(1) = 4 so perhaps try negative numbers as this result is getting bigger and we want to find y(a) = 0 for a being a factor

y(-1) = 0 BINGO x = -1 is a factor so we divide the above equation by x +1 to get a quadratic

Answer 5

x^2(2x + 1) + 1

about as far as you can go!