Question

Factorise 2 + x^3 - 3x^6

answer is (2 + 3x^3)(1 - x)(1 + x + x^2)
need working out

Answer 1

Let w = x^3

Square both sides to find that w^2 = (x^3)^2 = x^(3*2) = x^6

In short: w^2 = x^6

The given expression 2+x^3-3x^6 turns into 2+w-3w^2 and rearranges into -3w^2+w+2

Set this equal to zero and use the quadratic formula. We'll plug in

• a = -3
• b = 1
• c = 2

So,

`w = (-b\pm√(b^2-4ac))/(2a)\\\\w = (-1\pm√((1)^2-4(-3)(2)))/(2(-3))\\\\w = (-1\pm√(25))/(-6)\\\\w = (-1\pm5)/(-6)\\\\w = (-1+5)/(-6) \ \text{ or } \ w = (-1-5)/(-6)\\\\w = (4)/(-6) \ \text{ or } \ w = (-6)/(-6)\\\\w = -(2)/(3) \ \text{ or } \ w = 1\\\\`

If w = -2/3, then that rearranges to the following

w = -2/3

3w = -2

3w+2 = 0

This makes (3w+2) a factor of -3w^2+w+2

If w = 1, then it rearranges to w-1 = 0.

This makes (w-1) a factor of -3w^2+w+2

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To summarize the previous section, we found the factors of -3w^2+w+2 were:

• (3w+2)
• (w-1)

It leads to (3w+2)(w-1)

We must stick a negative out front because the leading coefficient is negative.

Therefore, -3w^2+w+2 = -(3w+2)(w-1)

You can use the FOIL rule to confirm.

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Recall we made w = x^3

Let's replace each w with x^3

-(3w+2)(w-1)

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

This tells us that 2+x^3-3x^6 factors to -(3x^3+2)(x^3-1)

The next task is to factor x^3-1 using the difference of cubes factoring rule.

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

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

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

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So,

2+x^3-3x^6

-3x^6 + x^3 + 2

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

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

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

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

Take careful notice that x-1 turned into -(1-x) in the 3rd step. The negative out front for -(1-x) cancels out with the original negative out front.