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1. State the remainder theorem: [1]
Either “dividend = divisor × quotient + remainder”
Or f(x) = (x – a)Q(x) + R
2. Solve for p, q and r: 3x2 + 6x – 4 ≡ p(x –q) 2 + r [3]
The left side is the expanded form. The right side is the vertex form.
We can convert the left to the right or from the right to the left. Right to left is probably easier.
right side = p(x –q) 2 + r
right side = p(x2 –2qx +q2) + r
right side = px2 –2pqx +pq2 + r
left side = 3x2 + 6x – 4
By comparison, 3x2 = px2 --> p = 3
By comparison, 6x = -2pqx = -2(3)qx = -6qx --> q = -1
By comparison, – 4 = pq2 + r --> r = -7
3. The polynomial P(x) = x3 + ax2 + bx – 3 has a remainder of 27 when divided by x – 2, and a remainder of 3 when divided by x + 1. Calculate the remainder when P(x) is divided by x – 1. [4]
i) x – 2 = 0 à x = 2
When x = 2, f(x) = 27 (Remainder theorem)
f(2) = 27 = 23 + a(2)2 + b(2) – 3
22 = 4a + 2b (Note all the coefficients are even)
11 = 2a + 1b (Equation 1)
ii) x +1 = 0 à x = -1
When x = -1, f(x) = 3 (Remainder theorem)
f(-1) = 3 = (-1)3 + a(-1)2 + b(-1) – 3
7 = 1a – 1b (Equation 2)
Eqn 1 + Eqn 2: 3a = 18 à a =6
Therefore b = -1
P(x) = x3 + 6x2 – x – 3
4. Is 2x – 3 a factor of 3x3 + 2x2 – 7x + 2? [2]
2x – 3 = 0
x = 1.5
3(1.5)3 + 2(1.5)2 – 7(1.5) + 2 ≠ 0
2x - 3 is not a factor.
5. Solve the equation 2x3 + 3x2 – 32x + 15 = 0. s09p1.q6 [6]
f(x) = (ax ± b)(1x ± c)(1x ± d)
Obviously a = 2.
b × c × d = 15
Candidates are ±1, ±3, ±5.
f(1) ≠ 0
f(-1) ≠ 0
f(3) = 54 +27 - 96 +15 = 0
(x – 3) is a factor.
f(5) ≠ 0
f(-5) = -250 + 75 + 160 + 15 = 0
(x - -5) = (x + 5) is a factor.
By inspection the third factor must be (2x – 1).
The solutions are x = -5, ½, 3.
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