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4 Polynomials

1
2

2

2

1

5

⫺1


1


1


3

2

6

2

So, from this we can say that
1
2x3 ⫹ x2 ⫹ 5x ⫺ 1 ⫽ ¢x ⫺ ≤12x2 ⫹ 2x ⫹ 62 ⫹ 2
2
⫽ 12x ⫺ 12 1 x2 ⫹ x ⫹ 32 ⫹ 2

Exercise 1
1 Evaluate f1x2 ⫽ x4 ⫺ 3x3 ⫹ 3x2 ⫹ 7x ⫺ 4 for x ⫽ 2.
2 Evaluate g1x2 ⫽ 7x3 ⫺ 2x2 ⫺ 8x ⫹ 1 for x ⫽ ⫺2.
3 Evaluate f1x2 ⫽ x6 ⫺ 4x3 ⫺ 7x ⫹ 9 for x ⫽ ⫺1.
4 Find f(4) for each polynomial.
a f 1x2 ⫽ x3 ⫺ x2 ⫹ 2x ⫺ 5

b f 1x2 ⫽ 5x4 ⫺ 4x2 ⫹ 8

c f 1t2 ⫽ t5 ⫺ 6t3 ⫹ 7t ⫹ 6

d f 1x2 ⫽ 6 ⫺ 7x ⫹ 5x2 ⫺ 2x3

˛

˛

˛

˛

1
5 Calculate f ¢⫺ ≤ for f1x2 ⫽ 6x3 ⫺ 4x2 ⫺ 2x ⫹ 3.
2
6 Use synthetic division to find the quotient and remainder for each of these
calculations.
b 1x3 ⫺ 4x2 ⫹ 5x ⫺ 12 ⫼ 1x ⫺ 12

a 1x2 ⫹ 6x ⫺ 32 ⫼ 1x ⫺ 22

c 12x3 ⫹ x2 ⫺ 8x ⫹ 72 ⫼ 1x ⫺ 62 d 1x3 ⫹ 5x2 ⫺ x ⫺ 92 ⫼ 1x ⫹ 4 2
e 1x4 ⫺ 5x2 ⫹ 3x ⫹ 72 ⫼ 1x ⫹ 12 f 1x5 ⫺ x2 ⫺ 5x ⫺ 112 ⫼ 12x ⫺ 12
g 1t3 ⫺ 7t ⫹ 92 ⫼ 12t ⫹ 12

h 1⫺3x4 ⫺ 4x3 ⫺ 5x2 ⫹ 132 ⫼ 14x ⫹ 32

7 Express each function in the form f1x2 ⫽ 1px ⫺ q2Q1x2 ⫹ R where Q(x) is the
quotient on dividing f(x) by 1px ⫺ q2 and R is the remainder.

90

f(x)

1px ⫺ q2

(a)

3x ⫺ 7x ⫹ 2

x⫺2

(b)

x3 ⫹ 6x2 ⫺ 8x ⫹ 7

x⫺5

(c)

4x ⫹ 7x ⫺ 9x ⫺ 17

x⫹3

(d)

5x5 ⫺ 4x3 ⫹ 3x ⫺ 2

x⫹4

(e)

2x ⫺ 5x ⫹ 9

x⫹1

(f)

x3 ⫺ 7x2 ⫹ 4x ⫺ 2

2x ⫺ 1

(g)

2x ⫺ 4x ⫹ 11

2x ⫹ 1

2

3

2

6

4

4

2

In this situation, there will
always be a common factor
in the quotient. This common
factor is the coefficient of x
in the divisor.