Question

A level maths factor theorem help in part (b) please​

Answer 1

Given polynomial $f(x)=4 x^{3}-12 x^{2}+2 x-6$.

(a) To show that (x – 3) is a factor of f(x).

Factor theorem:

A polynomial f(x) has a factor of (x – a), if and only if f(a) = 0.

$f(x)=4 x^{3}-12 x^{2}+2 x-6$

$f(3)=4 (3)^{3}-12 (3)^{2}+2 (3)-6$

       $=4(27)-12(9)+6-6$

       $=108-108$

       = 0

f(3) = 0

Hence (x – 3) is a factor of f(x).

(b) $f(x)=0$

$4 x^{3}-12 x^{2}+2 x-6=0$

Here $4x^2$ is common in first 2 terms and 2 is common in next 2 terms.

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

Now, (x – 3) is common in both terms.

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

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

x = 3 and $4x^2+2=0$

3 is the real root.

using quadratic formula solve $4x^2+2=0$ and find the other roots.

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$x=\frac{-0 \pm \sqrt{0^{2}-4 (4) (2)}}{2 (4)}$

$x=\pm\frac{\sqrt{-32}}{8}$

$x=\pm\frac{4\sqrt{2}i}{8}$

$x=\pm\frac{\sqrt{2}i}{2}$

These are complex roots.

Hence 3 is the only real root of the equation f(x) = 0.

Answer 2

Answer:

Quadratic has no real roots

Step-by-step explanation:

4x³ - 12x² + 2x - 6

4x²(x - 3) + 2(x - 3)

(4x² + 2)(x - 3)

The quadratic factor, 4x² + 2 has ni real roots because:

B² - 4AC

(0)² - 4(4)(2)

-32 < 0

So the only real root is from (x - 3), which is 3