Question

Show that $$(3x + 2)$$ is a factor of $$(6x^{3} + 13x^{2} - 4)$$ and hence factorise $$(6x^{3} + 13x^{2} - 4)$$.

Answer 1

Let $$f(x) = 6x^{3} + 13x^{2} - 4.$$ 

The factor $$(3x + 2)$$ is zero when $$x = -\frac{2}{3}.$$ 

By the Factor Theorem, we calculate

$$f\left(-\frac{2}{3}\right) = 6\left(-\frac{2}{3}\right)^{3} + 13\left(-\frac{2}{3}\right)^{2} - 4$$

$$= -\frac{16}{9} + \frac{52}{9} - \frac{36}{9}$$

$$= 0$$ 

Therefore, $$(3x + 2)$$ is a factor. 

On dividing the polynomial by $$(3x + 2)$$, we obtain $$6x^{3} + 13x^{2} - 4 = (3x + 2)(2x^{2} + 3x - 2).$$ 

Now, $$2x^{2} + 3x - 2 = 2x^{2} + 4x - x - 2$$

$$= 2x(x + 2) - (x + 2)$$

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

Hence, the complete factorisation is $${6x^{3} + 13x^{2} - 4 = (3x + 2)(2x - 1)(x + 2)}.$$