IB HL

www.ibmaths.com

Adrian Sparrow

Factor and remainder theorem

Factors and solutions

Factorise the quadratic x2-x-12. (x+3)(x-4)

Solve x2-x-12=0 x=-3 and x=4

Substitute x=-3 into the equation, and calculate the answer.

Repeat with x=4.

Both answers are 0, because these are factors.

Cubic polynomials

This technique will also work for higher order polynomials such as cubics.

Show that (x-3) is a factor x3-4x2+x+6.

Substitute in x=3. (3 3)-4(32)+3+6=0

Hence, using trial and error find the other 2 factors of x3-4x2+x+6.

Hint: these factors are usually integers, start with 0, then try 1, -1, 2, etc.

(x-2) (x+1)

Cubic polynomials - typical questions

Given that (x-2) and (x+1) are factors of the cubic polynomial 2x3+ax2+bx+6, find the values of a and b.

Substitute in the first factor x=2, make up a linear equation.

Substitute in the second factor x=(-1), make up another linear equation.

Solve simultaneously to find a and b.

2(23)+(22)a+2b+6=0

2(-13)+(-12)a-b+6=0

16+4a+2b+6=0

4a+2b=-22

-2+a-b+6=0

a-b=-4

a=-5, b=-1

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(x + 3) x 2 − x −12)

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x

Dividing by factors - quadratics

This technique is not used by the IB, but is useful to know and explains the remainders - division by an algebraic factor.

Divide the quadratic x2-x-12 by one of it’s factors, e.g. (x+3).

Divide x2 by x.

Take the answer and multiply by 3.

Take this away from the next component in the polynomial.

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−3x

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−4xDivide this answer by x.

Repeat the steps above.

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−4

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−12 -

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−12

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0Two things to note:

Remainder = 0The other factor is given as the answer.

Now repeat this process but divide the quadratic by (x-4)

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(x − 3) x 3 − 4x 2 + x + 6)

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−3x 2

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+3x

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−2x + 6

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+ 6

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0€

−x 2 + x

Dividing by factors - cubics

Divide the polynomial x3-4x2+x+6 by (x-3).

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x 2 − x −2

The quotient left as an answer can now be factorised to give the other two factors of the original cubic.

All 3 factors are:

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(x −2)(x +1)(x − 3)€

(x −2)(x +1)

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x 2 − x −2

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(x −2) x 3 − x 2 −10x −8)

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x 2 + x −8

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−2x 2

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x 2 −10x

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−2x

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−8x −8

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+16

Remainders when dividing

Divide the polynomial x3-x2-10x-8 by (x-2).

We have a remainder.

This tells us that (x-2) is not a factor of the polynomial.

Substitute x=2 into the polynomial x3-x2-10x-8.

Answer = -24€

−24

Using the remainder

Given that (x+1) is one of the factors of the cubic polynomial ax3+5x2+bx-9, and that when the polynomial is divided by (x-2) a remainder of 15 is left, find the values of a and b.

Substitute in the first factor x=-1, make up a linear equation.

Substitute in the value x=2, but the answer is 15, not 0. Make up a linear equation.

Solve simultaneously to find a and b.

a(-13)+5(-12)-b-9=0

a(23)+5(22)+2b-9=15

-a+5-b-9=0

a+b=-4

8a+20+2b-9=15

8a+2b=4

a=2, b=-6

Exam questions

2. Given that (x+1) is one of the factors of the cubic polynomial 3x3+ax2+bx+4, and that when the polynomial is divided by (x-1) a remainder of -12 is left, find the values of a and b.

1. Given that (x-2) and (x-5) are both factors of the cubic polynomial ax3+bx2-x+30, find the values of a and b.

3. a) Show that (x-3) is one of the factors of the cubic polynomial 2x3-3x2-14x+15.

a=2, b=-11

a=-10, b=-9

b) Hence find the remaining two factors of the cubic polynomial 2x3-3x2-14x+15.

2(33)-3(32)-14(3)+15=0

Hint: you will need to divide by (x-3).

(x-3)(x-1)(2x+5)