Question

If $$(x^3 + ax^2 + bx + 6)$$ has $$(x - 2)$$ as a factor and leaves a remainder 3 when divided by $$(x - 3)$$, find the values of $$a$$ and $$b$$.

Answer 1

Let $$f(x) = x^3 + ax^2 + bx + 6.$$

Since $$(x - 2)$$ is a factor, the Factor Theorem gives: $$f(2) = 0.$$

$$8 + 4a + 2b + 6 = 0$$

$$\Rightarrow 2a + b = -7 \quad (1)$$

By the Remainder Theorem, the remainder upon division by $$(x - 3)$$ is $$f(3)$$. 

Hence, $$f(3) = 3.$$ 

$$27 + 9a + 3b + 6 = 3$$

$$\Rightarrow 3a + b = -10 \quad{2}$$

Subtracting equation (1) from equation (2): $$a = -3.$$

Substituting in equation (1): $$2(-3) + b = -7 \Rightarrow b = -1.$$

Thus, $${a = -3, b = -1}.$$

With these values, $$f(x) = x^3 - 3x^2 - x + 6 = (x - 2)(x^2 - x - 3),$$ so the required factor and remainder conditions are satisfied.