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. Given also that (x + 1) is a factor of f(x), (b) find another equation in A and B. (c) Find the value of A and the value of B. (d) Hence find a quadratic factor of f(x).

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f(x) = 24x^3 + Ax^2 - 3x + B where A and B are constants - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 4

Question 4

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. Given also that (x + ... show full transcript

Worked Solution & Example Answer:f(x) = 24x^3 + Ax^2 - 3x + B where A and B are constants - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 4

Step 1

Show that A + 4B = 114.

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Answer

To show that A + 4B = 114, we need to use the remainder theorem. According to the theorem, if f(x) is divided by (2x - 1), the remainder can be obtained by substituting x = 1/2 into f(x).

Calculating f(1/2):

$f(1/2) = 24(1/2)^3 + A(1/2)^2 - 3(1/2) + B$

This simplifies to:

$f(1/2) = 24 \cdot \frac{1}{8} + A \cdot \frac{1}{4} - \frac{3}{2} + B$

Now, substituting the values leads to:

$f(1/2) = 3 + \frac{A}{4} - \frac{3}{2} + B$

We know the remainder is 30, hence:

$3 + \frac{A}{4} - \frac{3}{2} + B = 30$

Simplifying this, we get:

$\frac{A}{4} + B + 1.5 = 30$

Further simplifying yields:

$\frac{A}{4} + B = 28.5$

Multiplying everything by 4 to clear the fraction gives:

$A + 4B = 114.$

Thus, we have shown that A + 4B = 114.

Step 2

find another equation in A and B.

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Answer

Since (x + 1) is a factor of f(x), we substitute x = -1 into f(x):

$f(-1) = 24(-1)^3 + A(-1)^2 - 3(-1) + B = 0$

This simplifies to:

$-24 + A + 3 + B = 0$

Combining terms gives:

$A + B - 21 = 0$

Thus, we have the equation:

$A + B = 21.$

Step 3

Find the value of A and the value of B.

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Answer

We now have a system of two equations:

A + 4B = 114
A + B = 21

To solve for A and B, we can subtract the second equation from the first:

$(A + 4B) - (A + B) = 114 - 21$

This simplifies to:

$3B = 93$

Therefore:

$B = 31.$

Substituting B back into the second equation gives:

A = 21 - 31 \= -10.$$ Thus, the values are A = -10 and B = 31.

Step 4

Hence find a quadratic factor of f(x).

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Answer

Now that we have A and B, we can express f(x):

$f(x) = 24x^3 - 10x^2 - 3x + 31.$

To factorize f(x), we can use synthetic division with the factor (x + 1) to divide:

Perform synthetic division of f(x) by (x + 1).
The result of this division gives

$f(x) = (x + 1)(24x^2 - 34x + 31).$

Thus, the quadratic factor of f(x) is:

$24x^2 - 34x + 31.$

f(x) = 24x^3 + Ax^2 - 3x + B where A and B are constants - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 4

Worked Solution & Example Answer:f(x) = 24x^3 + Ax^2 - 3x + B where A and B are constants - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 4

Show that A + 4B = 114.

find another equation in A and B.

Find the value of A and the value of B.

Hence find a quadratic factor of f(x).

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