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 know (x + 3) is a factor, so we can perform polynomial long division to factor the function. Dividing:

$f(x) = -6x^3 - 7x^2 + 40x + 21$

by (x + 3) yields:

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

Next, we factor the quadratic -6x^2 + 11x + 7. We look for two numbers that multiply to -6 * 7 = -42 and add to 11. The numbers 14 and -3 work:

So, we can rewrite:

$-6x^2 + 14x - 3x + 7$

Grouping these gives:

$= -2x(3x - 7) - 1(3x - 7)$

Thus,

$= (x + 3)(-2x - 1)(3x - 7)$

This can be rewritten as:

$= -(x + 3)(2x + 1)(3x - 7)$

First, we simplify the left-hand side:

$6(2^3) + 7(2^2) = 6(8) + 7(4) = 48 + 28 = 76$

Now, we simplify the right-hand side:

$40(2) + 21 = 80 + 21 = 101$

Setting the equation up gives:

$76 = 101$

This does not hold, so we need to rearrange:

To find the value, let's rewrite the original equation and isolate variables:

$6(2^y) + 7(2^2) = 40(2) + 21$

Simplifying:

2^y = rac{101 - 28}{6} = rac{73}{6}

Taking logarithms

y = rac{ ext{log}(73/6)}{ ext{log}(2)}

Evaluating with a calculator gives:

$y ext{ approximately } 2.222$

Thus, the answer to two decimal places is:

$2.22$