f(x) = -6x^3 - 7x^2 + 40x + 21 (a) Use the factor theorem to show that (x + 3) is a factor of f(x) (b) Factorise f(x) completely - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 3

From part (a), we know that (x + 3) is a factor of f(x). We can divide f(x) by (x + 3) using either polynomial long division or synthetic division.

Performing synthetic division with -3:

 -3 | -6  0  -7  40  21  
    |    18  54  -3 -9  
    ------------------------- 
      -6  18  47  37  12  

This gives us: $f(x) = (x + 3)(-6x^2 + 18x + 7)$

Next, we need to factor the quadratic. We can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = -6$, $b = 18$, and $c = 7$.

Calculating:

$$\begin{align*} b^2 - 4ac & = 18^2 - 4(-6)(7) \ & = 324 + 168 \ & = 492, \ \end{align*}$$

Thus, the quadratic can be factored as follows: $f(x) = (x + 3)(-6)(x - (3 - \sqrt{41}))(x - (3 + \sqrt{41}))$

So, the complete factorization is:
$f(x) = -6(x + 3)(x - (3 - \sqrt{41}))(x - (3 + \sqrt{41}))$

First, we simplify the left and right sides of the equation:

Left side: $6(2^3) + 7(2^2) = 6(8) + 7(4) = 48 + 28 = 76.$

Right side: $40(2^1) + 21 = 40(2) + 21 = 80 + 21 = 101.$

Thus, we have the equation: $76 = 101,$
which is inconsistent. Therefore, no solutions exist for this equation.