How to factorise a cubic equation (Method 1) : ExamSolutions - Free Printable

We are asked to factorise the cubic expression:

$$
2x^3 - 3x^2 - 11x + 6
$$

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🔍 Step 1: Use the Rational Root Theorem

The Rational Root Theorem tells us that any rational root of the polynomial (i.e., a value of $ x $ that makes the expression zero) is of the form:

$$
\frac{p}{q}
$$

where:
- $ p $ is a factor of the constant term (here, $ 6 $),
- $ q $ is a factor of the leading coefficient (here, $ 2 $).

So:
- Factors of 6: $ \pm1, \pm2, \pm3, \pm6 $
- Factors of 2: $ \pm1, \pm2 $

Thus, possible rational roots are:

$$
\pm1, \pm2, \pm3, \pm6, \pm\frac{1}{2}, \pm\frac{3}{2}
$$

We will test these values in the polynomial until we find one that gives zero.

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🧪 Step 2: Test possible roots

Let’s define:

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

Try $ x = 1 $:

$$
f(1) = 2(1)^3 - 3(1)^2 - 11(1) + 6 = 2 - 3 - 11 + 6 = -6 \ne 0
$$

Try $ x = -1 $:

$$
f(-1) = 2(-1)^3 - 3(-1)^2 - 11(-1) + 6 = -2 - 3 + 11 + 6 = 12 \ne 0
$$

Try $ x = 2 $:

$$
f(2) = 2(8) - 3(4) - 11(2) + 6 = 16 - 12 - 22 + 6 = -12 \ne 0
$$

Try $ x = 3 $:

$$
f(3) = 2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0 ✔
$$

✔ So, $ x = 3 $ is a root, which means $ (x - 3) $ is a factor.

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🔁 Step 3: Polynomial Division or Synthetic Division

Now divide $ f(x) $ by $ (x - 3) $ to get a quadratic factor.

We’ll use synthetic division.

Set up synthetic division with root $ 3 $:

```
Coefficients: 2 -3 -11 6
| 6 9 -6
-------------------
2 3 -2 0
```

Explanation:
- Bring down 2.
- Multiply $ 2 \times 3 = 6 $, add to next: $ -3 + 6 = 3 $
- $ 3 \times 3 = 9 $, add to next: $ -11 + 9 = -2 $
- $ -2 \times 3 = -6 $, add to last: $ 6 + (-6) = 0 $

So the quotient is:

$$
2x^2 + 3x - 2
$$

Therefore:

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

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🔍 Step 4: Factor the quadratic

Now factor $ 2x^2 + 3x - 2 $

We look for two numbers that multiply to $ 2 \times (-2) = -4 $ and add to $ 3 $.

Try:
- $ 4 $ and $ -1 $: $ 4 \times (-1) = -4 $, $ 4 + (-1) = 3 $ ✔

So split the middle term:

$$
2x^2 + 4x - x - 2
$$

Group:

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

✔ So:

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

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✔ Final Answer:

$$
\boxed{(x - 3)(2x - 1)(x + 2)}
$$

This is the fully factored form of:

$$
2x^3 - 3x^2 - 11x + 6
$$

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✔️ Check (Optional)

Multiply back:

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

Then: $ (x - 3)(2x^2 + 3x - 2) $

$$
= x(2x^2 + 3x - 2) - 3(2x^2 + 3x - 2)
= 2x^3 + 3x^2 - 2x - 6x^2 - 9x + 6
= 2x^3 - 3x^2 - 11x + 6
$$

✔ Matches original!

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✔ Final Answer:

$$
\boxed{(x - 3)(2x - 1)(x + 2)}
$$