1. IB HL www.ibmaths.com Adrian Sparrow Factor and remainder theorem

2. Factors and solutions Factorise the quadratic x 2 -x-12.(x+3)(x-4) Solve x 2 -x-12=0x=-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.

3. Cubic polynomials This technique will also work for higher order polynomials such as cubics. Show that (x-3) is a factor x 3 -4x 2 +x+6. Substitute in x=3.(3 3 )-4(3 2 )+3+6=0 Hence, using trial and error find the other 2 factors of x 3 -4x 2 +x+6. Hint: these factors are usually integers, start with 0, then try 1, -1, 2, etc. (x-2) (x+1)

4. Cubic polynomials - typical questions Given that (x-2) and (x+1) are factors of the cubic polynomial 2x 3 +ax 2 +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(2 3 )+(2 2 )a+2b+6=0 2(-1 3 )+(-1 2 )a-b+6=0 16+4a+2b+6=0 4a+2b=-22 -2+a-b+6=0 a-b=-4 a=-5, b=-1

5. 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 x 2 -x-12 by one of it’s factors, e.g. (x+3). Divide x 2 by x. Take the answer and multiply by 3. Take this away from the next component in the polynomial. Divide this answer by x. Repeat the steps above. - Two things to note: Remainder = 0 The other factor is given as the answer. Now repeat this process but divide the quadratic by (x-4)

6. Dividing by factors - cubics Divide the polynomial x 3 -4x 2 +x+6 by (x-3). The quotient left as an answer can now be factorised to give the other two factors of the original cubic. All 3 factors are:

7. Remainders when dividing Divide the polynomial x 3 -x 2 -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 x 3 -x 2 -10x-8. Answer = -24

8. Using the remainder Given that (x+1) is one of the factors of the cubic polynomial ax 3 +5x 2 +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(-1 3 )+5(-1 2 )-b-9=0 a(2 3 )+5(2 2 )+2b-9=15 -a+5-b-9=0 a+b=-4 8a+20+2b-9=15 8a+2b=4 a=2, b=-6

9. Exam questions 2. Given that (x+1) is one of the factors of the cubic polynomial 3x 3 +ax 2 +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 ax 3 +bx 2 -x+30, find the values of a and b. 3. a) Show that (x-3) is one of the factors of the cubic polynomial 2x 3 -3x 2 -14x+15. a=2, b=-11 a=-10, b=-9 b) Hence find the remaining two factors of the cubic polynomial 2x 3 -3x 2 -14x+15. 2(3 3 )-3(3 2 )-14(3)+15=0 Hint: you will need to divide by (x-3). (x-3)(x-1)(2x+5)