A-Level Edexcel Maths Pure Integration: Let $f(x) = 2x^3

Let $f(x) = 2x^3 - 7x^2 + 4x + 4$ - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 1

Question 4

Let $f(x) = 2x^3 - 7x^2 + 4x + 4$. (a) Use the factor theorem to show that $(x - 2)$ is a factor of $f(x)$. (b) Factorise $f(x)$ completely.

Worked Solution & Example Answer:Let $f(x) = 2x^3 - 7x^2 + 4x + 4$ - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 1

Step 1

96%

114 rated

Answer

To use the factor theorem, we need to evaluate the function at the value of x that makes the factor zero. For $(x - 2)$ , we set $x = 2$ .

Calculate $f(2)$ :

$f(2) = 2(2)^3 - 7(2)^2 + 4(2) + 4$

Calculating each term:

So, we have: $f(2) = 16 - 28 + 8 + 4$ $f(2) = 0$

Since $f(2) = 0$ , by the factor theorem, $(x - 2)$ is indeed a factor of $f(x)$ .

Step 2

99%

104 rated

Answer

We begin by using synthetic division to divide $f(x)$ by $(x - 2)$ . The coefficients of $f(x)$ are [2, -7, 4, 4].

Setting up for synthetic division:

2 | 2   -7   4   4
   |      4 -6 -4
----------------------
   |  2   -3  -2  0

The remainder is 0, confirming that $(x - 2)$ is a factor. The quotient is $2x^2 - 3x - 2$ . Now, we need to factor this quadratic expression:

To factor $2x^2 - 3x - 2$ , we look for two numbers that multiply to $2 imes -2 = -4$ and add up to $-3$ . The numbers $-4$ and $1$ fit this.

Rewriting the quadratic, we have:

$2x^2 - 4x + x - 2$

Now we group the terms:

$=(2x^2 - 4x) + (x - 2)$

Factoring by grouping gives:

$= 2x(x - 2) + 1(x - 2)$

Factoring out the common factor $(x - 2)$ :

$= (x - 2)(2x + 1)$

Thus, the complete factorization of $f(x)$ is:

$f(x) = (x - 2)(2x + 1)(x - 2)$ or combined as $f(x) = (x - 2)^2(2x + 1)$ .