The expression f(x)=ax3+(a−3b)x2+3bx+cf(x) = ax^3 + (a - 3b)x^2 + 3bx + c is exactly divisible by x2+3xx^2 + 3x. When f(x)f(x) is divided by x+2x + 2, the remainder is 10. Find the values of aa, bb and cc. Hence factorise f(x)f(x) completely. Given that 3x3−2x2+x−4=A(x−1)+B(x−1)(x+1)+Cx(x2−1)+D3x^3 - 2x^2 + x - 4 = A(x - 1) + B(x - 1)(x + 1) + Cx(x^2 - 1) + D for all values of xx, find the value of AA, BB, CC and DD.

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Worked solution

8(a)

1. Use the fact that $f(x)$ is divisible by $x^2 + 3x$ to find $c$.

Since x2+3x=x(x+3)x^2 + 3x = x(x + 3), f(x)f(x) is divisible by xx and x+3x + 3.
By the Factor Theorem, f(0)=0f(0) = 0.
f(0)=a(0)3+(a−3b)(0)2+3b(0)+c=0f(0) = a(0)^3 + (a - 3b)(0)^2 + 3b(0) + c = 0
⇒c=0\Rightarrow c = 0

2. Use the Factor Theorem for $x + 3$ to form an equation in $a$ and $b$.

f(−3)=0f(-3) = 0
a(−3)3+(a−3b)(−3)2+3b(−3)+0=0a(-3)^3 + (a - 3b)(-3)^2 + 3b(-3) + 0 = 0
−27a+9(a−3b)−9b=0-27a + 9(a - 3b) - 9b = 0
−27a+9a−27b−9b=0-27a + 9a - 27b - 9b = 0
−18a−36b=0-18a - 36b = 0
a=−2ba = -2b --- (1)

3. Use the Remainder Theorem for $x + 2$ to form another equation in $a$ and $b$.

f(−2)=10f(-2) = 10
a(−2)3+(a−3b)(−2)2+3b(−2)+0=10a(-2)^3 + (a - 3b)(-2)^2 + 3b(-2) + 0 = 10
−8a+4(a−3b)−6b=10-8a + 4(a - 3b) - 6b = 10
−8a+4a−12b−6b=10-8a + 4a - 12b - 6b = 10
−4a−18b=10-4a - 18b = 10
−2a−9b=5-2a - 9b = 5 --- (2)

4. Solve the simultaneous equations to find $a$ and $b$.

Substitute (1) into (2):
−2(−2b)−9b=5-2(-2b) - 9b = 5
4b−9b=54b - 9b = 5
−5b=5⇒b=−1-5b = 5 \Rightarrow b = -1
Substitute b=−1b = -1 into (1):
a=−2(−1)=2a = -2(-1) = 2

5. Factorise $f(x)$ completely.

f(x)=2x3+(2−3(−1))x2+3(−1)x+0f(x) = 2x^3 + (2 - 3(-1))x^2 + 3(-1)x + 0
f(x)=2x3+5x2−3xf(x) = 2x^3 + 5x^2 - 3x
Since x2+3xx^2 + 3x is a factor, f(x)=(x2+3x)(2x−1)f(x) = (x^2 + 3x)(2x - 1)
f(x)=x(x+3)(2x−1)f(x) = x(x + 3)(2x - 1)

Answer

a=2a = 2, b=−1b = -1, c=0c = 0; f(x)=x(x+3)(2x−1)f(x) = x(x+3)(2x-1)

8(b)

1. Substitute $x = 1$ to find $D$.

Let x=1x = 1:
3(1)3−2(1)2+1−4=A(0)+B(0)+C(0)+D3(1)^3 - 2(1)^2 + 1 - 4 = A(0) + B(0) + C(0) + D
3−2+1−4=D3 - 2 + 1 - 4 = D
D=−2D = -2

2. Substitute $x = -1$ to find $A$.

Let x=−1x = -1:
3(−1)3−2(−1)2+(−1)−4=A(−1−1)+B(0)+C(0)+D3(-1)^3 - 2(-1)^2 + (-1) - 4 = A(-1 - 1) + B(0) + C(0) + D
−3−2−1−4=−2A+D-3 - 2 - 1 - 4 = -2A + D
−10=−2A−2-10 = -2A - 2
−2A=−8-2A = -8
A=4A = 4

3. Compare coefficients of $x^3$ to find $C$.

Comparing the coefficients of x3x^3 on both sides:
LHS coefficient of x3x^3 is 33.
RHS coefficient of x3x^3 comes from Cx(x2−1)=Cx3−CxCx(x^2 - 1) = Cx^3 - Cx, which is CC.
Therefore, C=3C = 3.

4. Substitute $x = 0$ to find $B$.

Let x=0x = 0:
3(0)3−2(0)2+0−4=A(−1)+B(−1)(1)+C(0)+D3(0)^3 - 2(0)^2 + 0 - 4 = A(-1) + B(-1)(1) + C(0) + D
−4=−A−B+D-4 = -A - B + D
−4=−4−B−2-4 = -4 - B - 2
B=−2B = -2

Answer

A=4A = 4, B=−2B = -2, C=3C = 3, D=−2D = -2

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