Page 29 - Pra U STPM 2022 Penggal 1 - Mathematics (T)

Mathematics Term 1 STPM Chapter 1 Functions

Example 19

4
3
2
1 Find the remainder when the polynomial 2x – 5x + x – 7x + 1 is divided by (2x + 1).
4
2
3
Solution: Let P(x) = 2x – 5x + x – 7x + 1
Since 2x + 1 is the divisor, we choose a value of x such that 2x + 1 = 0,
1
i.e. x = – .

2
1
1
1
1
1
P – 1 2 = 2 – 1 2 4 – 5 – 1 2 3 + – 1 2 2 – 7 – 1 2 + 1
2
2
2
2
2
= 1 + 5 + 1 + 7 + 1
8 8 4 2
= 11 or 5 1
2 2
Hence, the remainder when the polynomial 2x – 5x + x – 7x + 1 is divided by
4
2
3
1
(2x + 1) is 5 .
2
1
When a polynomial P(x) is divided by (ax + b), the remainder is P – b 2 .
a
Example 20
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3
2
The polynomial ax – 5x + bx – 7x + 1 leaves a remainder of –8 when it is divided by (x – 1), and a
remainder of 11 when divided by (2x + 1). Determine the values of a and b.
2
2
3
4
Solution: Let P(x) = ax – 5x + bx – 7x + 1
Hence, P(1) = –8
i.e. a – 5 + b – 7 + 1 = –8
a + b = 3 ………… 
1
Also, P – 1 2 = 11
2
2
1
1
1
1
i.e. a – 1 2 4 – 5 – 1 2 3 + b – 1 2 2 2 – 7 – 1 2 + 1 = 11
2
2
2
2
a + 5 + b + 7 + 1 = 11
16 8 4 2 2
a + 10 + 4b + 56 + 16 = 88
a + 4b = 6 ………… 
 – : 3b = 3
b = 1
Substitute b = 1 into :
a + 1 = 3
a = 2
Example 21
When the polynomial P(x) is divided by (x – 1), its remainder is 5. When P(x) is divided by (x – 2), its
remainder is 7. Given that P(x) may be written in the form (x – 1)(x – 2) · Q(x) + Ax + B, where Q(x) is
a polynomial, A and B are constants, find the remainder when P(x) is divided by (x – 1)(x – 2).
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01a STPM Math T T1.indd 26 3/28/18 4:20 PM