The diagram shows the graph of $y = x^{3} - 3x^{2} - x - 12$ .

(i)
Explain why $x^{3} - 3x^{2} - x - 12$ cannot be written as a product of 3 linear factors.
(ii)
Show that $x - 4$ is a factor of $x^{3} - 3x^{2} - x - 12$ .
(iii)
Factorise $x^{3} - 3x^{2} - x - 12$ completely and hence solve the equation

$x^{3} - 3x^{2} - x - 12 = 5(x - 4).$

Remainder factor theorems 中文

Work through it with AI →

It won't just hand you the answer — it asks where you're stuck.How the checker works

Worked solution

3(ii)

1. Evaluate the polynomial at $x = 4$ using the Factor Theorem

Let $P(x) = x^3 - 3x^2 - x - 12$ .
$P(4) = (4)^3 - 3(4)^2 - (4) - 12$
$P(4) = 64 - 48 - 4 - 12$
$P(4) = 0$

Answer

Since $P(4) = 0$ , $x - 4$ is a factor.

3(iii)

1. Find the quadratic factor of the cubic expression using the linear factor from part (ii)

By long division or comparing coefficients:
$x^3 - 3x^2 - x - 12 = (x - 4)(x^2 + bx + c)$
Comparing $x^2$ coefficients: $-4 + b = -3 \implies b = 1$
Comparing constants: $-4c = -12 \implies c = 3$
So, $x^3 - 3x^2 - x - 12 = (x - 4)(x^2 + x + 3)$

2. Substitute the factorised form into the equation and rearrange

$(x - 4)(x^2 + x + 3) = 5(x - 4)$
$(x - 4)(x^2 + x + 3) - 5(x - 4) = 0$
$(x - 4)(x^2 + x + 3 - 5) = 0$
$(x - 4)(x^2 + x - 2) = 0$

3. Factorise the resulting quadratic and solve for $x$

$(x - 4)(x + 2)(x - 1) = 0$
$x = 4$ , $x = -2$ , or $x = 1$

Answer

$(x - 4)(x^2 + x + 3)$
$x = -2, 1, 4$

AI analysis — may contain mistakes.

Worked this one out on paper? Snap your steps and Gigle's AI math checker will circle the step where it went wrong.

The diagram shows the graph of y=x3−3x2−x−12y = x^{3} - 3x^{2} - x - 12y=x3−3x2−x−12.

Worked solution

3(ii)

3(iii)

More questions on this topic

The diagram shows the graph of $y = x^{3} - 3x^{2} - x - 12$ .