Question

If (x – 2) is a factor of the expression \(2x^3 + ax^2 + bx - 14\) and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b. 

[Hint : Let \(p(x) = 2x^3 + ax^2 + bx - 14\). Then, \(p(2) = 0\) and \(p(3) = 52\).]

Answer 1

Let \(p(x) = 2x^{3} + ax^{2} + bx - 14\).

Since \((x - 2)\) is a factor, \(p(2) = 0\).

\(p(2) = 2(2)^{3} + a(2)^{2} + b(2) - 14\)

\(= 16 + 4a + 2b - 14\)

\(= 4a + 2b + 2\)

So, \(4a + 2b + 2 = 0\)

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

By Remainder Theorem, when \(p(x)\) is divided by \((x - 3)\), the remainder is \(p(3)\). 

Given \(p(3) = 52\).

\(p(3) = 2(3)^{3} + a(3)^{2} + b(3) - 14\)

\(= 54 + 9a + 3b - 14\)

\(= 9a + 3b + 40\)

So, \(9a + 3b + 40 = 52\)

\(\Rightarrow 9a + 3b = 12\)

\(\Rightarrow 3a + b = 4 \quad \ldots (2)\)

Subtracting (1) from (2):

\((3a + b) - (2a + b) = 4 - (-1)\)

\(\Rightarrow a = 5\)

Putting \(a = 5\) in (1):

\(2(5) + b = -1\)

\(\Rightarrow 10 + b = -1\)

\(\Rightarrow b = -11\)

Hence, \(a = 5\) and \(b = -11\).