WEEK 4
TOPIC: polynomial (continued)
CONTENT:
Factorization of polynomial
Show that x – 1 is a factor off(x) = 2x3 + 3x2 – 5x – 6
Solution
f(-1) = 2(-1)3 + 3(-1)2 – 5(-1) – 6
= -2 + 3 + 5 – 6
Hence x + 1 is a factor of f(x)
Show that x + 1 is a factor of f(x) = x3 + 2x2 – 5x – 6
Hence factorise f(x) completely.
Solution
f(-1) =(-1)3 + 2(-1)2 – 5(-1) – 6
= -1 + 2 + 5 – 6 = 0
: x+1 is a factor of f(x)
Using Long Division:
x2 + x – 5
(a) x + 1 x3 + 2x2 – 5x – 6
x3 + x2
x2 – 5x
x2 + x
-6x – 6
-6x – 6
0
x2 + x – 6 = x2 + 3x2 – 2x – 6
= x(x + 3) -2 (x + 3)
= (x + 3)(x – 2)
Hence f(x) = (x + 1)(x + 3) (x – 2)
Factiorizef(x) = x3 + 7x2 – 14x – 8 completely.
Solution
In a complete factorised form, f(x) can be written in the form:
f(x) = (x ± p)(x ± r) …(1)
The first term of the expansion of (1) is x3 while the last term of the expansion is ±pqr. Hence, p. q and r must be factors of -8. We can therefore try:
x ± 1
x ± 2
x ± 4
x ± 8
f(-1) = -1 – 7 – 14 – 8 = -30 ≠ 0
: x + r is not a factor of f(x)
f(1) = 1 – 7 + 14 – 8 = 0
: x – 1 is a factor of f(x).
A similar procedure can be used for x ± 2, x ± 4, x ± 8 to find the other two factors. But a less cumbersome procedure is to use long division method once a factor of f(x) is obtained.
x2 – 6x + 8
x – 1 x2 – 7x2 + 14x – 8
x2 –x2
-6x2 + 14x
-6x2+ 6x
8x – 8
8x – 8
0
Now, x2 – 6x + 8 = x2 – 4x – 2x + 8
= (x – 2)(x – 4)
Hence, f(x) = (x – 1)(x – 2)(x – 4)
What must be subtracted from f(x) = x2+2x2-3x +5 so that it will be exactly divisible by x – 2? Hence, find that polynomial H(x) which is exactly divisible by x – 2 and factorise it completely.
Solution
f(x) = x2 + 2x2 – 3x + 5
f(2) = 8 + 8 – 6 + 5
= 15
:When f(x) is divided by x – 2 the remainder is 15. Hence 15 must be subtracted from f(x) to make it exactly divisible by x – 2.
Let the new polynomial which is exactly divisible by x – 2 be H(x), then:
H(x) = f(x) – 15
= x2 + 2x2 – 3x + 5– 15
= x2 + 2x2 – 3x –10
Using long division
x2 + 4x +5
x– 2 x2 + 2x2 – 3x – 10
x2 + 2x2
4x2 – 3x
4x2 – 8x
5x – 10
5x – 10
0
But x2 + 4x + 5 is not factorizable, hence H(x) = (x – 2)(x2 + 4x + 5).
Given that f(x – 1) = x2 + 3x –1, find f(3).
Put x + 1 = 3
x = 2
: f(3) = (2)2 + 3(2) – 1
= 4 + 6 – 1
= 9
(1) Given that p1(x) = 5x3 + 3x2 – 2x + 6, p2(x) = x3 + 4x2 – 3x + 1 and p3(x) 2x3 – 3x + 2, find
(i) (p2 – p3) p1 (ii)p2(p1 + p3)
(2) Divide (i) 4x3 + 3x2 – 2x + 1 by x2 + 2x – 1 (ii) x4 + 2x + 3 by x2 – 1 to obtain the
remainder and quotient.
Evaluation
Find the quadratic equation whose roots are (i) 3 & -2 (ii) -1 & 8 (iii) ¾ & ½
General Evaluation
(1) When x2 + bx + 2 is divided by x + 3 the remainder is 5. Find the value of b.
(2) If 2x2 – (b – 4) x – 4 (b + 2) = 0, has equal roots, find the possible values of b.
(3) Factorise completely x3 + 5x2 – 3x + 1
(4) Solve the following pair of equation simultaneously 4x – 3y = 17, 3x2 – 2y2 + x – 4y = 73
Reading Assignment
Further Maths 1 pages 66 – 69 Exercise 6a Q5, 9 and 10
Weekend Assignment
(1) Find the remainder when 2x3 – 4x2 + x – 3 is divided by x + 3
(a) 84 (b) -86 (c) -64 (d) 76
(2) Factorise x4 – 1 completely (a) (x + 2) (x – 3) (b) (x + 1) (x – 1) (x2 + 1)
(c) (x + 1) (x + 2) (x – 3) (d) (x2 + 1) (x + 1) (x – 2)
(3) Given that p1(x) = 2x4 + 3x3 – x2 + 2x – 3 and p1(x) = 3x3 + 2x + 2. Find 3p1(x) – 3p1(x)
(a) 6x4 – 3x2 – 15 (b) 5x4 – 3x4 + 2x2 – x + 3 (c) 6x4 = 9x3 – 15
(d) 6x4 + 18x2 + 6x2
(4) Given that p(x) = x3 + 4x2 – 3x + 1, find p( ½ )
(a)
(5) Given that p(x) = ax2 + bx + 1, p( ½ ) = ½ and p(-2) = 23, determine the values of a and b
(a) a = -3, b = 4 (b) a = 2, b = -3 (c) a = 4, b = -3 (d) a = 3, b = -3
Theory
(1) Find the quotient and remainder when 2x4 – 3x3 + x2 – 4x + 5 is divided by x2 + 3x + 1
(2) If p1 = 3x3 + 2x2 – x + 2, p2 = 2x2 + x – 6 and p3 = x3 + 3x2 + 2x – 4, find
(i) p3(p1 + p2) (ii) p2 + p3 – 3p1 (iii) p2 x (p3 + p1)