Question

If x3 + ax2 + bx + 6 has (x – 2) as a factor and leaves a remainder 3 when divided by (x – 3),
find the values of a and b.

Answer 1

Answer:

The value of a and b are -3 and -1 respectively.

Step-by-step explanation:

Given that (x-2) is a factor of polynomial

$P(x)=x^3+ax^2+bx+6$

Also when divided by (x - 3) leaves a remainder 3.

we have to find the value of a and b.

As 2 is the zero of the polynomial therefore by remainder theorem

$P(2)=0$

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

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

$4a+2b+14=0$

$2a+b=-7$     →   (1)

$\text{Also the polynomial }x^3+ax^2+bx+6\text{ when divided by (x - 3) leaves a remainder 3}$

∴ $P(3)=3$

$3^3+a(3)^2+3b+6=3$

$27+9a+3b+6=3$

$9a+3b+33=3$

$3a+b=-10$     →   (2)

Solving (1) and (2), we get

Subtracting equation (2) from (1)

$2a+b-3a-b=-7-(-10)$

$-a=3$

$a=-3$

$2a+b=-7$

$2(-3)+b=-7$

$b=-7+6=-1$

Hence, the value of a and b are -3 and -1 respectively.

Answer 2

Answer:

$Value \: of \: a = -3 \\b=-1$

Step-by-step explanation:

Let p(x)=x³+ax²+bx+6

i ) it is given that , (x-2) is a factor of p(x) ,then

p(x)=0

=> 2³+a(2)²+b×2+6=0

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

=> 4a+2b+14=0

Divide each term by 2 , we get

=> 2a+b+7=0 ---(1)

ii)It is given that, if p(x) divided by (x-3) leaves a remainder 3

=> p(3)=3

=> 3³+a(3)²+b×3+6=3

=> 27+9a+3b+6-3=0

=> 9a+3b+30=0

Divide each term by 3 , we get

=> 3a+b+10=0 ---(2)

Subtract equation (1) from equation (2) , we get

a + 3 = 0

=> a = -3

Now,

Substitute a=-3 in equation (1) , we get

2(-3)+b+7=0

=> -6+b+7=0

=> b+1=0

=> b = -1

Therefore,

$Value \: of \: a = -3 \\b=-1$

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