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Parents Pricing Home A-Level Edexcel Maths Pure Circles f(x) = 24x^3 + Ax^2 - 3x + B
where A and B are constants
f(x) = 24x^3 + Ax^2 - 3x + B
where A and B are constants - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 4 Question 5
View full question f(x) = 24x^3 + Ax^2 - 3x + B
where A and B are constants.
When f(x) is divided by (2x - 1) the remainder is 30
(a) Show that A + 4B = 114
(2)
Given also that (x... show full transcript
View marking scheme Worked Solution & Example Answer:f(x) = 24x^3 + Ax^2 - 3x + B
where A and B are constants - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 4
Show that A + 4B = 114 Only available for registered users.
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To find the remainder when dividing by (2x - 1), use the Remainder Theorem, which states that the remainder of f(x) when divided by (ax + b) can be found by evaluating f(-b/a). In this case, substituting x = 1/2 into f(x):
R = f ( 1 / 2 ) = 24 ( 1 / 2 ) 3 + A ( 1 / 2 ) 2 − 3 ( 1 / 2 ) + B R = f(1/2) = 24(1/2)^3 + A(1/2)^2 - 3(1/2) + B R = f ( 1/2 ) = 24 ( 1/2 ) 3 + A ( 1/2 ) 2 − 3 ( 1/2 ) + B Calculating each term:
24 ( 1 / 2 ) 3 = 24 ( 1 / 8 ) = 3 24(1/2)^3 = 24(1/8) = 3 24 ( 1/2 ) 3 = 24 ( 1/8 ) = 3 A ( 1 / 2 ) 2 = A ( 1 / 4 ) = A 4 A(1/2)^2 = A(1/4) = \frac{A}{4} A ( 1/2 ) 2 = A ( 1/4 ) = 4 A − 3 ( 1 / 2 ) = − 3 2 -3(1/2) = -\frac{3}{2} − 3 ( 1/2 ) = − 2 3
Setting up the equation:
3 + A 4 − 3 2 + B = 30 3 + \frac{A}{4} - \frac{3}{2} + B = 30 3 + 4 A − 2 3 + B = 30 Simplifying:
A 4 + B + 3 − 3 2 = 30 \frac{A}{4} + B + 3 - \frac{3}{2} = 30 4 A + B + 3 − 2 3 = 30 Bringing all terms together:
A 4 + B + 6 2 − 3 2 = 30 A 4 + B + 3 2 = 30 A 4 + B = 30 − 3 2 A 4 + B = 60 2 − 3 2 = 57 2 \frac{A}{4} + B + \frac{6}{2} - \frac{3}{2} = 30
\frac{A}{4} + B + \frac{3}{2} = 30
\frac{A}{4} + B = 30 - \frac{3}{2}
\frac{A}{4} + B = \frac{60}{2} - \frac{3}{2} = \frac{57}{2} 4 A + B + 2 6 − 2 3 = 30 4 A + B + 2 3 = 30 4 A + B = 30 − 2 3 4 A + B = 2 60 − 2 3 = 2 57 Multiplying through by 4 for clarity gives:
A + 4 B = 114 A + 4B = 114 A + 4 B = 114
find another equation in A and B. Only available for registered users.
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Since (x + 1) is a factor of f(x), by the Factor Theorem, we have:
f ( − 1 ) = 0 f(-1) = 0 f ( − 1 ) = 0 Substituting x = -1 into f(x):
f ( − 1 ) = 24 ( − 1 ) 3 + A ( − 1 ) 2 − 3 ( − 1 ) + B = 0 ⟹ − 24 + A + 3 + B = 0 f(-1) = 24(-1)^3 + A(-1)^2 - 3(-1) + B = 0 \implies -24 + A + 3 + B = 0 f ( − 1 ) = 24 ( − 1 ) 3 + A ( − 1 ) 2 − 3 ( − 1 ) + B = 0 ⟹ − 24 + A + 3 + B = 0 This simplifies to:
A + B − 21 = 0 ⟹ A + B = 21 A + B - 21 = 0 \implies A + B = 21 A + B − 21 = 0 ⟹ A + B = 21
Find the value of A and the value of B. Only available for registered users.
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With the equations:
A + 4 B = 114 A + 4B = 114 A + 4 B = 114 A + B = 21 A + B = 21 A + B = 21
We can solve for A and B.
( A + 4 B ) − ( A + B ) = 114 − 21 (A + 4B) - (A + B) = 114 - 21 ( A + 4 B ) − ( A + B ) = 114 − 21 This simplifies to:
3 B = 93 ⟹ B = 31 3B = 93 \implies B = 31 3 B = 93 ⟹ B = 31 Substituting B back into the second equation:
A + 31 = 21 ⟹ A = 21 − 31 = − 10 A + 31 = 21 \implies A = 21 - 31 = -10 A + 31 = 21 ⟹ A = 21 − 31 = − 10
Hence find a quadratic factor of f(x). Only available for registered users.
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To find a quadratic factor using the found values of A and B:
Substituting A and B into f(x), we have:
f ( x ) = 24 x 3 − 10 x 2 − 3 x + 31 f(x) = 24x^3 - 10x^2 - 3x + 31 f ( x ) = 24 x 3 − 10 x 2 − 3 x + 31 We know (x + 1) is a factor, so we use polynomial long division or synthetic division to divide:
Calculating gives us:
f ( x ) = ( x + 1 ) ( 24 x 2 − 34 x + 31 ) f(x) = (x + 1)(24x^2 - 34x + 31) f ( x ) = ( x + 1 ) ( 24 x 2 − 34 x + 31 ) So, the quadratic factor is:
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