1.

The expression 2x3 + ax2 + bx – 2 leaves a remainder 7 and 0 when divided by (2x – 3) and (x + 2). Find the values of a and b. A) a = 3, b = -3 B) a = -3, b = -3 C) a = -3, b = 3 D) a = 3, b = 3

Answer»

Correct option is (A) a = 3, b = -3

Let \(p(x)=2x^3 + ax^2 + bx – 2\)

\(\because\) p(x) leaves a remainder 7 when divided by (2x – 3).

\(\therefore\) \(p(\frac32)=7\)

\(\Rightarrow\) \(2(\frac32)^3+a(\frac32)^2+b\times(\frac32)–2=7\)

\(\Rightarrow\) \(\frac{27}4+\frac{9a}4+\frac{3b}2-2=7\)

\(\Rightarrow\) 27+9a+6b-8 = 28

\(\Rightarrow\) 9a+6b = 9

\(\Rightarrow\) 3a+2b = 3   __________(1)

Also, p(x) leaves remainder 0 when divided by (x+2).

\(\therefore\) p(-2) = 0

\(\Rightarrow2(-2)^3+a(-2)^2+b\times-2-2=0\)

\(\Rightarrow\) -16 + 4a - 2b - 2 = 0

\(\Rightarrow\) 4a - 2b = 18  __________(2)

Adding (1) & (2), we get

7a = 21 \(\Rightarrow\) \(a=\frac{21}7=3\)

\(\therefore\) 2b = 3 - 3a = 3 - 9 = -6 \(\Rightarrow\) \(b=\frac{-6}2=-3\)

Thus, a = 3, b = -3

Correct option is A) a = 3, b = -3



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