Find “a” and “b” in order that x3 – 6x2 + ax + b may be exa

1.	Find “a” and “b” in order that x3 – 6x2 + ax + b may be exactly divisible by x2 – 3x + 2. A) a = 11, b = -5 B) a = 11, b = 6 C) a = 11, b = -6 D) a = -11, b = -5
Answer» Correct option is (C) a = 11, b = -6 \((x^2 – 3x + 2)\) is a factor of \((x^3-6x^2+ax+b)\) \(\Rightarrow\) (x-1) & (x-2) are factors of \((x^3-6x^2+ax+b)\) \((\because\) (x-1) (x-2) \(=x^2 – 3x + 2)\) \(\Rightarrow\) x = 1 & x = 2 are zeros of \((x^3-6x^2+ax+b)\) \(\therefore\) 1 - 6 + a + b = 0 and 8 - 24 + 2a + b = 0 \(\Rightarrow\) a+b = 5 and 2a+b = 16 \(\Rightarrow\) (2a+b) - (a+b) = 16 - 5 \(\Rightarrow\) a = 11 \(\therefore\) b = 5 - a = 5 - 11 = -6 Correct option is B) a = 11, b = 6

Find “a” and “b” in order that x3 – 6x2 + ax + b may be exactly divisible by x2 – 3x + 2. A) a = 11, b = -5 B) a = 11, b = 6 C) a = 11, b = -6 D) a = -11, b = -5

Answer»

Correct option is (C) a = 11, b = -6

\((x^2 – 3x + 2)\) is a factor of \((x^3-6x^2+ax+b)\)

\(\Rightarrow\) (x-1) & (x-2) are factors of \((x^3-6x^2+ax+b)\)

\((\because\) (x-1) (x-2) \(=x^2 – 3x + 2)\)

\(\Rightarrow\) x = 1 & x = 2 are zeros of \((x^3-6x^2+ax+b)\)

\(\therefore\) 1 - 6 + a + b = 0 and 8 - 24 + 2a + b = 0

\(\Rightarrow\) a+b = 5 and 2a+b = 16

\(\Rightarrow\) (2a+b) - (a+b) = 16 - 5

\(\Rightarrow\) a = 11

\(\therefore\) b = 5 - a = 5 - 11 = -6

Correct option is B) a = 11, b = 6