Answer:
Hey There!
Let's solve....
First let's use the remainder theorem...
[tex]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 [/tex]
[tex]4a + 2b + 6 + 8 = 0 \\ 4a + 2b + 14 = 0 \\ so \: equation \: number \: 1 \: is \\ \boxed{2a + b + 7 = 0}[/tex]
Now it is given that...
[tex] {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 [/tex]
[tex]3a + b + 10 = 0 \\ \\ it \: is \: equation \: number \: 2[/tex]
So...
Let's find out the value of a
[tex]3a + \cancel{b} + 10 = 0 \\ 2a + \cancel{b} + 7 = 0 \\ a + 3 = 0 \\ \\ value \: of \: a \: is \\ \boxed{a = - 3}[/tex]
Let's find value of b...
[tex]2a + b + 7 = 0 \\ 2x - 3 + b + 7 = 0 \\ b + 1 = 0 \\ value \: of \: b \: is \\ \boxed{b = - 1}[/tex]
So a= -3 and b= -1 is the final answer .....
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
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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