Cambridge Additional Mathematics

172 Polynomials (Chapter 6)

If the leading coefficient of the polynomial 6 =1, then we need to multiply by this as well:

a(x¡®)(x¡ ̄)(x¡°)=ax^3 ¡a(®+ ̄+°)x^2 +a(® ̄+ ̄°+°®)x¡a® ̄°.

If you think a cubic equation has integer roots, try to find them by factorising the constant term.

Example 20 Self Tutor

Solve forx: x^3 ¡ 31 x¡30 = 0.

Let P(x)=x^3 ¡ 31 x¡ 30.

The constant term is¡ 30 , so the product of the roots is 30.

Since 30 = 5£ 3 £ 2 £ 1 , likely integer roots are § 1 ,§ 2 ,§ 3 ,§ 5. They could also be§ 6 since

2 £3=6, and so on.

Now P(1) =¡ 60 ,so 1 is not a root.

But P(¡1) = 0,so¡ 1 is a root, and (x+1)is a factor of P(x).

So, P(x)=x^3 +0x^2 ¡ 31 x¡30 = (x+ 1)(x^2 +bx¡30)

=x^3 +(b+1)x^2 +(b¡30)x¡ 30

Equatingx^2 s: b+1=0

) b=¡ 1

Hence P(x)=(x+ 1)(x^2 ¡x¡30)

=(x+ 1)(x+ 5)(x¡6)

) the solutions are¡ 1 ,¡ 5 , and 6.

Note that this method only works for those cubics with all integer roots.

EXERCISE 6E

1 Solve forx:

a x^3 ¡ 6 x^2 +11x¡6=0 b x^3 ¡ 3 x^2 +4=0 c x^3 +2x^2 ¡x¡2=0

d x^3 ¡ 6 x^2 +5x+12=0 e x^3 +5x^2 ¡ 16 x¡80 = 0 f x^3 +13x^2 +55x+75=0

2 Solve forx:

a 2 x^3 ¡ 6 x^2 ¡ 8 x+24=0 b 2 x^3 ¡ 2 x^2 ¡ 48 x¡72 = 0 c 3 x^3 ¡ 24 x^2 ¡ 15 x+ 252 = 0

The coefficient of

x^3 is 1 £1=1

The constant term

is 1 £¡30 =¡ 30

Take out a common

factor first!

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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_06\172CamAdd_06.cdr Friday, 20 December 2013 1:01:16 PM BRIAN