C2 Question on factorising

Tillybop

18

How do you factorise x³ + x² - 5x - 2 = 0

Thanks in advance :hat2:

Reply 1

TenOfThem

18

Original post by Tilly-Elizabeth

How do you factorise x³ + x² - 5x - 2 = 0

Thanks in advance :hat2:

The usual way is to "spot" a factor and the use one of the division methods to find the quadratic

Since the constant is -2 the factors to consider would be 1, -1, 2, -2

Have you used the factor or remainder theorem

Reply 2

Tillybop

OP

18

Original post by TenOfThem

The usual way is to "spot" a factor and the use one of the division methods to find the quadratic

Since the constant is -2 the factors to consider would be 1, -1, 2, -2

Have you used the factor or remainder theorem

(x-2) is a factor of the equation and when I use the factor theorem it substitutes perfectly. Ohhh it just clicked what I have to do.

One second I'm going to work it out and I'll see if I have it right.

Thanks :h:

Reply 3

TenOfThem

18

Original post by Tilly-Elizabeth

(x-2) is a factor of the equation and when I use the factor theorem it substitutes perfectly. Ohhh it just clicked what I have to do.

One second I'm going to work it out and I'll see if I have it right.

Thanks :h:

ok

Reply 4

Valyrian

18

Oh my god you're so cute Tilly :h:

Reply 5

Tillybop

OP

18

Original post by TenOfThem

ok

I have confused myself.

I have (x-2) which is one of the brackets so I'll have (x-2)(x-?)(x-?) = 0

So how do I remove the other two brackets?

Reply 6

TenOfThem

18

Original post by Tilly-Elizabeth

I have confused myself.

I have (x-2) which is one of the brackets so I'll have (x-2)(x-?)(x-?) = 0

So how do I remove the other two brackets?

x^3 + x^2 - 5x - 2 = (x - 2)(x^2 + bx + 1)

You can either use inspection or multiply out the RHS and compare coefficients

Then, when you have your quadratic you can check to see if that factorises

Reply 7

Tillybop

OP

18

Original post by TenOfThem

x^3 + x^2 - 5x - 2 = (x - 2)(x^2 + bx + 1)

You can either use inspection or multiply out the RHS and compare coefficients

Then, when you have your quadratic you can check to see if that factorises

Ah right.

So if I use algebraic division I get x² + 3x + 1

If I use the formula I get x=(-3-√5)/2 and x=(-3+√5)/2

So I should get:

(x-2)(x + 2.62)(x + 0.38) = 0

Reply 8

TenOfThem

18

Original post by Tilly-Elizabeth

Ah right.

So if I use algebraic division I get x² + 3x + 1

If I use the formula I get x=(-3-√5)/2 and x=(-3+√5)/2

So I should get:

(x-2)(x + 2.62)(x + 0.38) = 0

It depends on what the question asks

If you want to factorise then the answer is

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

If it asks you to solve

x^3 + x^2 - 5x - 2 = 0

Then your solutions are x = 2 or

x = \dfrac{-3\pm\sqrt5}{2}

Reply 9

Tillybop

OP

18

Original post by TenOfThem

It depends on what the question asks

If you want to factorise then the answer is

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

If it asks you to solve

x^3 + x^2 - 5x - 2 = 0

Then your solutions are x = 2 or

x = \dfrac{-3\pm\sqrt5}{2}

Oh I read it wrong it wanted the solutions.

Thank you for your help! :smile:

Reply 10

AstroNandos

21

Have you heard of the remainder theorem? This is what my teacher taught us for factorising polynomials. You can Google it if this website doesn't help, it's only one of many that exist :tongue:

http://www.purplemath.com/modules/remaindr.htm

(edited 9 years ago)

Reply 11

Zen-Ali

8

Few ways of doing this. Say

r

is a root, then by the factor theorem

(x - r)

is a factor. Equivalently,

f(r) = 0.

After you've guessed one, you can compare coefficients, get a quadratic in

x

(has to be a quadratic as the product of a linear and quadratic term is a cubic term), and then factorise (it's not necessarily factorisable) to get three linear terms.