Page 28 - Pra U STPM 2022 Penggal 1 - Mathematics (T)

Mathematics Term 1 STPM Chapter 1 Functions

6. Find, using long division, the quotient and remainder for
3
2
(a) (3x – 2x + 5) ÷ (x + 1) (b) (4x – 4x + 5x + 1) ÷ (2x – 3)
2
(c) (x – 3) ÷ (x + 3) (d) (3x – x – 6x + 11x – 1) ÷ (3x – 1)
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3
4
5
3
4
(e) (x – 2x + 6x – 5) ÷ (x – x – 1) 1
2
3
2
3
2
7. The expression ax – 8x + bx + 6 is divisible by x – 2x – 3. Find the values of a and b.
8. Find the values of A and B such that
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x + 4x + 4x + 1 ≡ (x + 1)(x + Ax + B)
2
3
9. Find J(x) if
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x + 3x + x – 2 ≡ (x + 2) · J(x)
2
10. Find numbers a, b and c, such that
2x + 2x + 5x + 3x + 3 ≡ (x + x + 1)(ax + bx + c)
4
2
2
3
2
for all values of x.
The remainder theorem
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When the polynomial 2x – 7x + 11x – 7 is divided by (x – 2), the quotient is 2x – 3x + 5 with remainder 3.
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2
We can write
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3
2x – 7x + 11x – 7 2 3 Remainder
––––––––––––––––– = 2x – 3x + 5 + –––––
x – 2 14243 x – 2
Quotient

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2
or 2x – 7x + 11x – 7 = (2x – 3x + 5)(x – 2) + 3 Remainder (A constant)
3
14424443 14243123
The dividend Quotient Divisor
(Polynomial (Polynomial (Polynomial
degree 3) degree 2) degree 1)

In general, if a polynomial of degree n, P(x), is divided by (x – a), the quotient, Q(x), is a polynomial of degree
(n – 1), and the remainder, R, is a constant, i.e.
P(x)
R
x – a = Q(x) + x – a
or P(x) ≡ Q(x) · (x – a) + R
By substituting x = a, we see that P(a) = R.
When a polynomial P(x) is divided by (x – a), the remainder is P(a).
Example 18
2
Find the remainder when the polynomial P(x) = 2x + 7x – 5x – 4 is divided by (x + 3).
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Solution: P(x) = 2x + 7x – 5x – 4
2
3
Since x + 3 is the divisor, choose a value of x such that x + 3 = 0, i.e. x = –3.
3
2
P(–3) = 2(–3) + 7(–3) – 5(–3) – 4
= –54 + 63 + 15 – 4
= 20
Hence, the remainder when the polynomial P(x) is divided by (x + 3) is 20.
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