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 8 - 2017 - Paper 3

We can start by dividing f(x) by (x + 3). Performing polynomial long division:

1. Divide the leading term: $-6x^3 / x = -6x^2$.
2. Multiply: $-6x^2 (x + 3) = -6x^3 - 18x^2$.
3. Subtract: $(-7x^2 + 18x^2) = 11x^2$.
4. Bring down the next term: $11x^2 + 40x$.
5. Divide $11x^2 / x = 11x$.
6. Multiply: $11x(x + 3) = 11x^2 + 33x$.
7. Subtract: $(40x - 33x) = 7x$.
8. Bring down the constant: $7x + 21$.
9. Finally, divide $7x / x = 7$.
10. Multiply: $7(x + 3) = 7x + 21$.
11. Subtract: $21 - 21 = 0$.

The complete factorisation of f(x) is:

$f(x) = (x + 3)(-6x^2 + 11x + 7)$

Now, we can factorise the quadratic: Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-11 \pm \sqrt{11^2 - 4(-6)(7)}}{-12}$

Calculating:

$x = \frac{-11 \pm \sqrt{121 + 168}}{-12} = \frac{-11 \pm \sqrt{289}}{-12} = \frac{-11 \pm 17}{-12}$

This gives:

$x = \frac{6}{-12} = -\frac{1}{2} \quad (1) \\ x = \frac{-28}{-12} = \frac{7}{3} \quad (2)$

Thus, the complete factorisation is:

$f(x) = -(x + 3)(x + \frac{1}{2})(x - \frac{7}{3})$.

We first replace the powers:

$6(8) + 7(4) = 40(2) + 21$

Calculating:

1. $6(8) = 48$
2. $7(4) = 28$
3. $40(2) = 80$

So the equation becomes:

$48 + 28 = 80 + 21$

Now compute the left side:

$76 = 101$

This is incorrect, so we solve for x:

The original equation is now manipulated into:

$6x^3 + 7x^2 = 40x + 21$

Substituting values found from previous factorisation, we see:

the values of x are:\n- $-3$\n- $-\frac{1}{2}$\n- $\frac{7}{3}$