If (x ^ 3 + a * x ^ 2 + bx + 6) has (x  2) as a factor and leaves a remainder 3 when divided by (x3), find the values of a and b
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1. User Answers ItsYourDeepak254
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 .....
I hope it is helpful to you..
Cheers!__________

2. User Answers vintechnology
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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