Johnny and Mary write a test on polynomials.

1. Complete the following statement about the factor theorem:

If f(b) = 0, ……………………………………..

Johnny’s answer: “If f(b) = 0, (x – b) is a factor of f(x)”

Mary’s answer: “If f(b) = 0, f(x) = (x –b) × Q(x)”

Both are correct.

2a) The remainder when the expression

*x*^{3}– 11*x*^{2}+*kx*– 30 is divided by*x*– 1 is 4 times the remainder when this expression is divided by*x*– 2. Find the value of the constant*k*.Johnny uses the Remainder Theorem.

The remainder when f(x) is divided by

*x*– 2 = f(2)r = f(2)

r = 8 – 44 + 2k – 30

r = 2k – 66

Johnny knows two facts about division by (x – 1):

The remainder when f(x) is divided by

*x*– 1 = f(1) The remainder when f(x) is divided by

*x*– 1 = 4 × f(2) = 4r 4r = f(1)

4r = 1 – 11 + 1k – 30

4r = k – 40

Johnny substitutes for 4r

2k – 66 = k – 40

k = 32

Mary uses long division. She gets the same answer, although her method is not as efficient.

b) Solve the equation

*x*^{3}– 4*x*^{2}– 8*x*+ 8 = 0, expressing non-integer solutions in the form*a*± √*b*, where*a*and*b*are integers.By trial and error, Mary finds P(-2) = 8 – 16 +16 + 8 = 0

Therefore (x –

**-**2) = (x + 2) is a linear factor.Mary reasons that a cubic equation must have three factors, but reference in the question to “

*a*± √*b”*is a very obvious hint that she will need to use the quadratic formula. Hence, she sees no point in continuing to look for other factors by trial and error!Mary uses Polynomial Identity to find the quadratic factor.

P(x) =

*x*^{3}– 4*x*^{2}– 8*x*+ 8 = (x + 2) (a*x*^{2}+b*x*+ c)1

*x*^{3}– 4*x*^{2}– 8*x*+ 8 = x(a*x*^{2}+b*x*+ c) + 2(a*x*^{2}+b*x*+ c)1

*x*^{3}– 4*x*^{2}– 8*x*+ 8 = a*x*^{3}+b*x*^{2}*+ cx + 2a**x*^{2}+2b*x*+ 2c1

*x*^{3}– 4*x*^{2}– 8*x*+ 8 = a*x*^{3}+(b +2a)*x*^{2}*+ (c +2b)x**+ 2*Mary knows when two polynomials are identical the coefficient of each term (degree by degree) must be identical.

She equates the third degree terms, 1

*x*^{3}= a*x*^{3}a = 1

She equates the constant (or degree zero) terms, 8 = 2c

c = 4

She substitutes a = 1 and c = 4 into (b +2a)

*x*^{2}*+ (c +2b)x*Therefore,

*b = -6*and the quadratic factor is 1*x*^{2}– 6*x*+ 4.Now Mary uses the quadratic formula.

Her solutions for P(x) = 0 are x = 3 + √5, or x = 3 √5, or x= 2. Excellent work!

Notes: Question 2 is taken from Cambridge International Education IGCSE Additional Mathematics

**winter 07 paper 1 question 8.**
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