Page 31 - Pra U STPM 2022 Penggal 1 - Mathematics (T)

Mathematics Term 1 STPM Chapter 1 Functions

If x = 2, P(2) = 8 – 14 – 6
= –12 ≠ 0
1 Hence, (x – 2) is not a factor of P(x).
If x = –2, P(–2) = (–2) – 7(–2) – 6
3
= –8 + 14 – 6
= 0
Hence, (x + 2) is a factor of P(x).
If x = 3, P(3) = 3 – 7(3) – 6
3
= 27 – 21 – 6
= 0
Hence, (x – 3) is a factor of P(x).
Since P(x) is of degree 3, it has only three linear factors.
3
P(x) = x – 7x – 6
= (x + 1)(x + 2)(x – 3)

Apart from the above method of using trial and error to obtain all the factors of a polynomial, we can also use
long division method, as shown in Example 23 below.

Example 23

Show that (x + 2) is a factor of f(x) = 6x + 13x – 4.
3
2
Hence, factorise f(x) completely and find the values of x such that f(x) = 0.
3
2
Solution: f(x) = 6x + 13x – 4
2
3
f(–2) = 6(–2) + 13(–2) – 4

= –48 + 52 – 4
= 0
Hence, by factor theorem, (x + 2) is a factor of f(x).
Using long division,
2
6x + x – 2
3
2
x + 2 6x + 13x + 0x – 4
3
2
6x + 12x
x + 0x
2
2
x + 2x
– 2x – 4
– 2x – 4
0
2
f(x) = (x + 2)(6x + x – 2)
= (x + 2)(3x + 2)(2x – 1)
When x = –2, f(–2) = 0
2
1
When x = – , f – 2 2 = 0
3 3
1
When x = 1 , f 1 2 = 0
2 2
2
The values of x such that f(x) = 0 are –2, – and 1 .
3
2
2
Note that –2, – and 1 are called the zeros of f.
3 2
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01a STPM Math T T1.indd 28 3/28/18 4:20 PM