Let us solve the given problem step by step.

The problem:

1. $P(x) = 2x^3 + x^2 - 13x + 6$
   • (i) Find all rational zeros of $P(x) = 0$.
   • (ii) Write $P(x)$ in factored form (hint: using synthetic division).

Step 1: Rational Root Theorem

To find rational zeros, the Rational Root Theorem suggests that any rational zero is of the form: $\pm \frac{p}{q}$ where:

• $p$: factors of the constant term (6),
• $q$: factors of the leading coefficient (2).

Factors of 6: $\pm 1, \pm 2, \pm 3, \pm 6$
Factors of 2: $\pm 1, \pm 2$

Possible rational zeros are: $\pm 1, \pm \frac{1}{2}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm 6$

Step 2: Test Possible Rational Roots

We test these values in $P(x) = 2x^3 + x^2 - 13x + 6$.

Testing $x = 1$:

$P(1) = 2(1)^3 + (1)^2 - 13(1) + 6 = 2 + 1 - 13 + 6 = -4 \quad \text{(Not a root)}.$

Testing $x = -1$:

$P(-1) = 2(-1)^3 + (-1)^2 - 13(-1) + 6 = -2 + 1 + 13 + 6 = 18 \quad \text{(Not a root)}.$

Testing $x = 2$:

$P(2) = 2(2)^3 + (2)^2 - 13(2) + 6 = 16 + 4 - 26 + 6 = 0 \quad \text{(Root found!)}.$

So, $x = 2$ is a root.

Step 3: Synthetic Division

Divide $P(x)$ by $x - 2$ using synthetic division.

Synthetic Division Setup: