Photo AI

1. (a) Simplify \( \frac{3x^3 - x - 2}{x^2 - 1} \) - Edexcel - A-Level Maths Pure - Question 3 - 2006 - Paper 4

Question icon

Question 3

1.-(a)-Simplify-\(-\frac{3x^3---x---2}{x^2---1}-\)-Edexcel-A-Level Maths Pure-Question 3-2006-Paper 4.png

1. (a) Simplify \( \frac{3x^3 - x - 2}{x^2 - 1} \). (b) Hence, or otherwise, express \( \frac{3x^3 - x - 2}{x^2 - 1} \) as a single fraction in its simplest form.

Worked Solution & Example Answer:1. (a) Simplify \( \frac{3x^3 - x - 2}{x^2 - 1} \) - Edexcel - A-Level Maths Pure - Question 3 - 2006 - Paper 4

Step 1

Simplify \( \frac{3x^3 - x - 2}{x^2 - 1} \)

96%

114 rated

Answer

To simplify ( \frac{3x^3 - x - 2}{x^2 - 1} ), we first factor the denominator. The denominator can be factored as:

x21=(x1)(x+1)x^2 - 1 = (x - 1)(x + 1)

Next, we attempt to factor the numerator, (3x^3 - x - 2). We can use polynomial long division or synthetic division to simplify this and find:

3x3x2=(3x+2)(x1)3x^3 - x - 2 = (3x + 2)(x - 1)

Therefore, substituting back, we have:

( \frac{(3x + 2)(x - 1)}{(x - 1)(x + 1)} )

Cancelling (x - 1) from the numerator and denominator gives:

3x+2x+1\frac{3x + 2}{x + 1}

Step 2

Express \( \frac{3x^3 - x - 2}{x^2 - 1} \) as a single fraction

99%

104 rated

Answer

We start with the previous result:

3x+2x+1\frac{3x + 2}{x + 1}

We also have to add the term ( \frac{1}{x(x + 1)} ). To combine these into a single fraction, we find a common denominator, which will be (x(x + 1)). Thus, re-writing the first term:

(3x+2)xx(x+1)=3x2+2xx(x+1)\frac{(3x + 2)x}{x(x + 1)} = \frac{3x^2 + 2x}{x(x + 1)}

Now, we can express the overall fraction as:

3x2+2x1x(x+1)\frac{3x^2 + 2x - 1}{x(x + 1)}

To simplify further, we check for common factors or any possible cancellations. In the numerator, we can factor:

3x2+2x1=(3x1)(x+1)3x^2 + 2x - 1 = (3x - 1)(x + 1)

Thus, combining this, we obtain:

(3x1)(x+1)x(x+1)\frac{(3x - 1)(x + 1)}{x(x + 1)}

Finally, we cancel the (x + 1):

3x1x\frac{3x - 1}{x}

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;