Question

please answer the question.The polynomial ax³+bx²+x-6 has(x+2) as a factor and leaves a remainder when divided by(x-2).Find a and b.​

Answer 1

Step-by-step explanation:

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Answer 2

${} \bf{ \huge{ \ \boxed{ \underline{ \mathfrak{ \blue{Answer:-}}}}}}$

=>a=0 and b=2.

${} \bf{ \huge{ \</p><p>\boxed{ \underline{ \mathfrak{ \blue{ Explanation:-}}}}}}$

Let p(x) =ax³+bx²+x-6.

By using factor theorem,(x-2) can be a factor of p(x) only when p(-2)=0.

p(-2)=a(-2)³+b(-2)²+(-2)-6=0.

=>-8a +4b-8=0.

Therefore,-2a+b =2...(1)

Also,when p(x) is divided by (x-2) the remainder is 4.

Therefore,p(2)=4.

=>a(2)³+b(2)²+2-6=4

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

=>8a+4b-4=4

=>8a+4b=4+4

=>8a+4b=8

=>2a+b=2...(2)

Adding equations (1) and (2),we get (-2a+b)+(2a+b)=2+2.

=>2b=4

$= > \frac{2b}{2} = \frac{4}{2} \: dividing \: both \: sides \: by \: 2$

=>b=2.

Putting b=2 in (1) ,we get

-2a +2=2

=>-2a=2-2

=>-2a=0

$=> \frac{( - 2a)}{( - 2) } = \frac{0}{( - 2)} dividing \: both \: sides \: by( - 2)$

=>a=0.

$\boxed{Hence ,\: a = 0 \: and \: b = 2}$