Question

If $$(2x^3 + ax^2 + bx - 2)$$ when divided by $$(2x - 3)$$ and $$(x + 3)$$ leaves remainders 7 and –20 respectively, find the values of $$a$$ and $$b$$.

Answer 1

Let $$f(x) = 2x^3 + ax^2 + bx - 2$$.

The value that makes $$2x - 3 = 0$$ is $$x = \frac{3}{2}$$.

$$f\left(\frac{3}{2}\right) = \frac{27}{4} + \frac{9a}{4} + \frac{3b}{2} - 2 = 7.$$

Multiplying by $$4$$,

$$27 + 9a + 6b - 8 = 28$$

$$\Rightarrow 9a + 6b = 9$$

$$\Rightarrow 3a + 2b = 3$$

For the divisor $$x + 3$$, the remainder is

$$f(-3) = 9a - 3b - 56 = -20$$

$$\Rightarrow 9a - 3b = 36$$

$$\Rightarrow 3a - b = 12$$

Thus,

$$3a + 2b = 3,$$

$$3a - b = 12.$$

Subtracting gives $$3b = -9$$, so $$b = -3$$.

Substituting in $$3a - b = 12$$ gives

$$3a + 3 = 12$$

$$\Rightarrow a = 3$$

Hence, $$a = 3$$ and $$b = -3$$.