Question

If (x ^ 3 + a * x ^ 2 + 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

The values of a and b are -3 and -1 respectively

x³ + ax ²+ bx + 6 has (x - 2) as a factor. Therefore,

(2)³ + a2²+ 2b + 6 = 0

8 + 4a + 2b + 6 = 0

4a + 2b = -14

x³ + ax ²+ bx + 6 leaves a remainder 3 when divided by (x - 3). Therefore,

(3)³ + a3²+ 3b + 6 = 3

27 + 9a + 3b + 6 = 3

9a + 3b = 3 - 33

9a + 3b = -30

Simultaneous equation:

4a + 2b = -14

9a + 3b = -30

2b = -14 -4a

b = -7 - 2a

Therefore,

9a + 3( -7 - 2a) = -30

9a - 21 - 6a = -30

3a = -30 + 21

3a = -9

a = -9 / 3

a = -3

4a + 2b = -14

4(-3) + 2b = -14

-12 + 2b = -14

2b = -14 + 12

2b = -2

b = -2 / 2

b = -1

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Answer 2

Answer:

Hey There!

Let's solve....

First let's use the remainder theorem...

$put \: x - 2 = 0 \\ \boxed{x = 2} \\ \\ f(x) = {x}^{3} + {ax}^{2} + bx + 6 \\ \\ f(2) = 0 = {2}^{3} + {2}^{2}a + 2b \\ + 6 = 0$

$4a + 2b + 6 + 8 = 0 \\ 4a + 2b + 14 = 0 \\ so \: equation \: number \: 1 \: is \\ \boxed{2a + b + 7 = 0}$

Now it is given that...

${x}^{3} + {ax}^{2} + bx + 6 \\ f(3) \to \: remainder = 3 \\ f(3) = 3 \\ {3}^{3} + {3}^{2}a + 3b + 6 = 3 \\ 27 + 9a + 3b + 6 = 3 \\ 9a + 3b + 30 = 0$

$3a + b + 10 = 0 \\ \\ it \: is \: equation \: number \: 2$

So...

Let's find out the value of a

$3a + \cancel{b} + 10 = 0 \\ 2a + \cancel{b} + 7 = 0 \\ a + 3 = 0 \\ \\ value \: of \: a \: is \\ \boxed{a = - 3}$

Let's find value of b...

$2a + b + 7 = 0 \\ 2x - 3 + b + 7 = 0 \\ b + 1 = 0 \\ value \: of \: b \: is \\ \boxed{b = - 1}$

So a= -3 and b= -1 is the final answer .....

I hope it is helpful to you..

Cheers!__________