Question

The expression 2x³ + ax² + bx - 2 leaves the remainder 7 and 0 when divided by (2x - 3) and (x + 2) respectively calculate the value of a and b. With these value of a and b factorise the expression completely.

Answer 1

Let P(x) = 2x³ + ax² + bx - 2
when P(x) is divided by 2x - 3
`"P"(3/2) = 2(3/2)^3 + a(3/2)^2 + b(3/2)-2` = 7
= `(27)/(4) + (9)/(4)a + (3)/(2)b -2` = 7
= `(27 + 9a + 6b - 8)/(4)` = 7
= 9a + 6b = 28 + 8 - 27
= 9a + 6b = 9
⇒ 3a + 2b = 3                            ...(1)
Similarly when P(x) is divided by x + 2
x = -2
2(-2)³ + a(-2)² + b(-2) -2 = 0
-16 + 4a - 2b - 2 = 0
⇒ 4a - 2b = 18                           ...(2)
On Solving equation (1) and (2)
3a + 2b = 3
4a - 2b = 18
       7a = 21
         a = 3
On substituting value of a in equation (1)
3 x 3 + 2b = 3
2b = 3 - 9
b = `(-6)/(2)`
= -3
b = -3
a = 3, b = -3
On substituting value of a and b
2x³ + 3a² - 3x - 2
When x + 2 is a factor___
x + 2) 2x³ + 3x² - 3x - 2 (2x² - x - 1
         2x³ + 4x²
          -        -________
               - x² - 3x
               - x² - 2x
                +     +
                  -x -2
                  -x - 2
                 +     +
                     x
2x² - x - 1
= 2x² - 2x + x - 1
= 2x(x - 1) + 1(x - 1)
(x - 1) (2x + 1)
Hence required factors are
(x - 1) (x + 2) (2x + 1).