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Nitella [24]
3 years ago
14

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

Mathematics
2 answers:
Rina8888 [55]3 years ago
8 0

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:

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

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.

notka56 [123]3 years ago
4 0

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

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