Question

${ax}^{3} + {bx}^{2} + x - 6 \: has \: (x + 2) \: as \: a \: factor \: and \: leaves \: remainder \: 4 \: when \: divided \: by \: x - 2. \: find \: the \: value \: of \: a \: and \: b$
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Answer 1

$\mathfrak {\large {\underline {\underline {Aloha !!}}}}$

Question :- ax^3 + bx^2+x-6 has (x+2) as a factor and leaves remainder 4 when divided by (x-2) .Find the value of "a" and "b"

solution :-

given 2 cases are

(1) x + 2 is a factor

(2) x -2 leaves 4 as remainder

case (1)

since x + 2 is a factor it leaves remainder zero

x +2 = 0

x = -2

Substute x = (-2) in the given polynomial

ax^3 + bx^2+x-6 =0

a (-2)^3 + b (-2)^2 +(-2)-6 =0

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

-8a+4b -8 =0

-4 ( 2a -b +2)=0

2a -b +2 =0----------(1)

case (2)

x -2 =0

x =2

ax^3 + bx^2+x-6 =4

a (2)^3+b (2)^2+2-6 =4

8a +4b -4 = 4

8a +4b -8 =0

4 (2a +b -2) =0

2a + b -2 =0----------(2)

from (1) and (2)

2a -b +2 =0

2a +b -2 =0

(-) (-) (+)

-----------------

-2b +4 =0

-----------------

-2b = -4

b =2

Now Substute "b " value in eq (1)

2a - b +2 =0

2a -2+2=0

2a =0

a =0

》 a = 0 , b =2

verification :-

o (x^3)+ 2 (x^2)+ x -6

2x^2 + x -6

(x +2 ) as a factor , x = -2

2 (-2)^2 -2 -6=0

2× 4 -8

8 -8 =0

Hence verified.

(x -2) leaves remainder 4 ; x =2

2x^2 + x -6

2 (2)^2 + 2 -6

2 × 4 -4

8 -4 = 4

Hence verified.