Given that the expression 2x3+ax2+bx+c2x^3 + ax^2 + bx + c leaves the same remainder when divided by (x−5)(x - 5) and when divided by (2x+1)(2x + 1), show that 9a+2b=−919a + 2b = -91. Given that the remainder is −6-6 when the expression 2x3+ax2+bx+c2x^3 + ax^2 + bx + c is divided by (x−1)(x - 1), and (x+1)(x + 1) is a factor, calculate the value of cc.

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Worked solution

1(a)

1. Define the polynomial and apply the Remainder Theorem for division by $(x - 5)$.

Let f(x)=2x3+ax2+bx+cf(x) = 2x^3 + ax^2 + bx + c.
When divided by (x−5)(x - 5), the remainder is f(5)f(5).
f(5)=2(5)3+a(5)2+b(5)+c=250+25a+5b+cf(5) = 2(5)^3 + a(5)^2 + b(5) + c = 250 + 25a + 5b + c

2. Apply the Remainder Theorem for division by $(2x + 1)$.

When divided by (2x+1)(2x + 1), the remainder is f(−12)f(-\frac{1}{2}).
f(−12)=2(−12)3+a(−12)2+b(−12)+cf(-\frac{1}{2}) = 2(-\frac{1}{2})^3 + a(-\frac{1}{2})^2 + b(-\frac{1}{2}) + c
=2(−18)+14a−12b+c= 2(-\frac{1}{8}) + \frac{1}{4}a - \frac{1}{2}b + c
=−14+14a−12b+c= -\frac{1}{4} + \frac{1}{4}a - \frac{1}{2}b + c

3. Equate the two remainders and simplify to show the required equation.

Since the remainders are the same:
250+25a+5b+c=−14+14a−12b+c250 + 25a + 5b + c = -\frac{1}{4} + \frac{1}{4}a - \frac{1}{2}b + c
250+25a+5b=−14+14a−12b250 + 25a + 5b = -\frac{1}{4} + \frac{1}{4}a - \frac{1}{2}b
Multiply by 4 to clear denominators:
1000+100a+20b=−1+a−2b1000 + 100a + 20b = -1 + a - 2b
100a−a+20b+2b=−1−1000100a - a + 20b + 2b = -1 - 1000
99a+22b=−100199a + 22b = -1001
Divide by 11:
9a+2b=−919a + 2b = -91 (Shown)

Answer

Shown.

1(b)

1. Apply the Remainder Theorem for division by $(x - 1)$.

Let f(x)=2x3+ax2+bx+cf(x) = 2x^3 + ax^2 + bx + c.
Since the remainder is −6-6 when divided by (x−1)(x - 1):
f(1)=−6f(1) = -6
2(1)3+a(1)2+b(1)+c=−62(1)^3 + a(1)^2 + b(1) + c = -6
2+a+b+c=−62 + a + b + c = -6
a+b+c=−8— (1)a + b + c = -8 \quad \text{--- (1)}

2. Apply the Factor Theorem for $(x + 1)$.

Since (x+1)(x + 1) is a factor:
f(−1)=0f(-1) = 0
2(−1)3+a(−1)2+b(−1)+c=02(-1)^3 + a(-1)^2 + b(-1) + c = 0
−2+a−b+c=0-2 + a - b + c = 0
a−b+c=2— (2)a - b + c = 2 \quad \text{--- (2)}

3. Solve the simultaneous equations to find $b$.

Subtracting equation (2) from equation (1):
(a+b+c)−(a−b+c)=−8−2(a + b + c) - (a - b + c) = -8 - 2
2b=−102b = -10
b=−5b = -5

4. Use the result from part (a) to find $a$, and subsequently $c$.

From part (a), we have 9a+2b=−919a + 2b = -91.
Substitute b=−5b = -5:
9a+2(−5)=−919a + 2(-5) = -91
9a−10=−919a - 10 = -91
9a=−819a = -81
a=−9a = -9

Substitute a=−9a = -9 and b=−5b = -5 into equation (1):
−9−5+c=−8-9 - 5 + c = -8
−14+c=−8-14 + c = -8
c=6c = 6

Answer

c=6c = 6

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