Page 30 - Pra U STPM 2022 Penggal 1 - Mathematics (T)

Mathematics Term 1 STPM Chapter 1 Functions

Solution: Given that P(x) ≡ (x – 1)(x – 2) · Q(x) + Ax + B
When P(x) is divided by (x – 1), the remainder is 5.
Hence, P(1) = 5 1
i.e. 0 + A + B = 5
A + B = 5 ………… 
When P(x) is divided by (x – 2), the remainder is 7.
Hence, P(2) = 7
i.e. 0 + 2A + B = 7
2A + B = 7 ………… 
 – : A = 2
Substituting A = 2 into :
2 + B = 5
B = 3
Hence, P(x) ≡ (x – 1)(x – 2) · Q(x) + 2x + 3.
P(x) 2x + 3
and –––––––––––– = Q(x) + ––––––––––––
(x – 1)(x – 2) (x – 1)(x – 2)
Hence, when P(x) is divided by (x – 1)(x – 2), the remainder is (2x + 3).

Note: Example 21 above shows that when a polynomial P(x) is divided by a quadratic expression (x – a)(x – b),
the remainder is a linear function in the form Ax + B, where A and B are constants.

The factor theorem
From the remainder theorem, we have shown that when a polynomial P(x) is divided by (x – a), its remainder,
R, is P(a). On the other hand, if (x – a) is a factor of P(x), then its remainder is zero, i.e. R = P(a) = 0.
For a polynomial P(x), (x – a) is a factor of P(x) if and only if P(a) = 0.

This theorem is very useful in the factorisation of polynomials of degrees higher than 2. By using this theorem,
the linear factors of the polynomial may be obtained.

Example 22

Factorise P(x) = x – 7x – 6.
3
Solution: [Hints to obtain the factors:
3
Notice that the constant term for P(x) = x – 7x – 6 is –6.
Hence, if (x – a) were to be a factor of P(x), then a must be a factor of –6,

i.e. a = ±1, ±2, ±3 or ±6.
Try substituting these values into P(x) such that P(a) = 0.]
If x = 1, P(1) = 1 – 7 – 6
= –12 ≠ 0
Hence, (x – 1) is not a factor of P(x).

3
If x = –1, P(–1) = (–1) – 7(–1) – 6
= –1 + 7 – 6
= 0
Hence, (x + 1) is a factor of P(x).

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