if( x+2) is a factor of p(x)=ax^3+bx^2+x-6 and p(x) when divided by (x-2) leaves a reminder 4, prove that a=0 and b=2
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(x+2) is a factor of p(x)=ax³ + bx² + x - 6
Remainder = 4
Prove that, a=0 and b=2
(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0 So, p(-2) = a(-2)³ + b(-2)² - 2 - 6 = 0
-8a + 4b - 8 = 0
-8a + 4b = 8
-2a + b = 2
-2a - 2 + b = 0 -------------(1)
Now, if p(x) is divided by (x -2) then it leaves remainder 4.
So, p(2) = a(2)³ + b(2)² + 2 - 6 = 4
8a + 4b - 4 = 4
8a + 4b = 8
2a + b = 2
2a + 2 + 2a = 2
4a = 0 -------------(2)
Solving equations (1) and (2),
i) -2a + b - 2 = 0
-2a + b = 2
b = 2
ii) 4a = 0
a = 0
Hence, proved.
Remainder = 4
Prove that, a=0 and b=2
(x + 2) is the factor of p(x) , and we know this is possible only when p(-2) = 0 So, p(-2) = a(-2)³ + b(-2)² - 2 - 6 = 0
-8a + 4b - 8 = 0
-8a + 4b = 8
-2a + b = 2
-2a - 2 + b = 0 -------------(1)
Now, if p(x) is divided by (x -2) then it leaves remainder 4.
So, p(2) = a(2)³ + b(2)² + 2 - 6 = 4
8a + 4b - 4 = 4
8a + 4b = 8
2a + b = 2
2a + 2 + 2a = 2
4a = 0 -------------(2)
Solving equations (1) and (2),
i) -2a + b - 2 = 0
-2a + b = 2
b = 2
ii) 4a = 0
a = 0
Hence, proved.
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