A-Level Edexcel Maths Pure Transformations of Functions: Given that \[ \frac{3x^4

Given that \[ \frac{3x^4 - 2x^3 - 5x^2 - 4}{x^2 - 4} \equiv ax^2 + bx + c + \frac{dx + e}{x^2 - 4}, \quad x \neq \pm 2 \] find the values of the constants a, b, c, d and e. - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 7

Question 2

$Given-that-\[-\frac{3x^4---2x^3---5x^2---4}{x^2---4}-\equiv-ax^2-+-bx-+-c-+-\frac{dx-+-e}{x^2---4},-\quad-x-\neq-\pm-2-\]-find-the-values-of-the-constants-a,-b,-c,-d-and-e.-Edexcel-A-Level Maths Pure-Question 2-2013-Paper 7.png$

Given that \[ \frac{3x^4 - 2x^3 - 5x^2 - 4}{x^2 - 4} \equiv ax^2 + bx + c + \frac{dx + e}{x^2 - 4}, \quad x \neq \pm 2 \] find the values of the constants a, b, c, d... show full transcript

Worked Solution & Example Answer:Given that \[ \frac{3x^4 - 2x^3 - 5x^2 - 4}{x^2 - 4} \equiv ax^2 + bx + c + \frac{dx + e}{x^2 - 4}, \quad x \neq \pm 2 \] find the values of the constants a, b, c, d and e. - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 7

Step 1

Using long division by $ x^2 - 4 $

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Answer

We start by performing long division of ( 3x^4 - 2x^3 - 5x^2 - 4 ) by ( x^2 - 4 ). In the first step, divide the leading term: ( 3x^4 \div x^2 = 3x^2 ).

Next, multiply back: ( 3x^2 \cdot (x^2 - 4) = 3x^4 - 12x^2 ).

Subtracting this from the original polynomial gives: [ (3x^4 - 2x^3 - 5x^2 - 4) - (3x^4 - 12x^2) = -2x^3 + 7x^2 - 4. ]

Step 2

Continuing the division

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Now, we divide the new leading term: ( -2x^3 \div x^2 = -2x. )

Multiply back: ( -2x \cdot (x^2 - 4) = -2x^3 + 8x. )

Subtract again: [ (-2x^3 + 7x^2 - 4) - (-2x^3 + 8x) = 7x^2 - 8x - 4. ]

Step 3

Final steps of the division

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Next, divide the leading term again: ( 7x^2 \div x^2 = 7 ).

Multiply back: ( 7 \cdot (x^2 - 4) = 7x^2 - 28. )

Subtract: [ (7x^2 - 8x - 4) - (7x^2 - 28) = -8x + 24. ]

This produces a remainder of ( -8x + 24 ), leading to: [ \frac{-8x + 24}{x^2 - 4}. ]

Step 4

Final expression and constants

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We can now express the left-hand side as: [ 3x^2 - 2x + 7 + \frac{-8x + 24}{x^2 - 4} ] Comparing coefficients:

From ( ax^2 ), we have ( a = 3 ).
From ( bx ), we have ( b = -2 ).
The constant term gives us ( c = 7 ).
From the numerator ( dx + e ), we have ( d = -8 ) and ( e = 24 ).

Thus, ( a = 3, b = -2, c = 7, d = -8, e = 24 ).

Given that \[ \frac{3x^4 - 2x^3 - 5x^2 - 4}{x^2 - 4} \equiv ax^2 + bx + c + \frac{dx + e}{x^2 - 4}, \quad x \neq \pm 2 \] find the values of the constants a, b, c, d and e. - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 7

Worked Solution & Example Answer:Given that \[ \frac{3x^4 - 2x^3 - 5x^2 - 4}{x^2 - 4} \equiv ax^2 + bx + c + \frac{dx + e}{x^2 - 4}, \quad x \neq \pm 2 \] find the values of the constants a, b, c, d and e. - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 7

Using long division by \( x^2 - 4 \)

Continuing the division

Final steps of the division

Final expression and constants

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