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If ax^3+bx^2+x-6 has (x+2) as a factor and leaves a remainder 4 when divided by (x-2),find the value of 'a' and 'b'. ANSWER a=0 b=2?
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If ax^3+bx^2+x-6 has (x+2) as a factor and leaves a remainder 4 when d...
Let p(x)= ax³+bx²+x-6........Since, x+2 is a factor of p(x) . therefore, x= -2 is zero of p(x)= (-2)³a + b(-2)² + (-2)-6.....= -8a+4b-8(#eq.1).........Nxt, p(2)= 4 .....a(2)³+b(2)²+2-6=4...=8a+4b-4=4...8a+4b = 8 .....(dividing by 4 both sides) 2a+b= 2...= b= 2-2a... (putting value of b in #eq.1....-8a+4(2-2a)-8=0 .....-8a+8-8a-8 = 0 ....-16a=0....a=0/-16....,a=0.....Since,b = 2-2*0..=2-0=2..........therefore, a=0 & b=2..
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If ax^3+bx^2+x-6 has (x+2) as a factor and leaves a remainder 4 when d...
Given Polynomial and Conditions
The polynomial in question is:
\[ P(x) = ax^3 + bx^2 + x - 6 \]
It is given that:
- \( (x + 2) \) is a factor of \( P(x) \).
- \( P(2) = 4 \) (the polynomial leaves a remainder of 4 when divided by \( (x - 2) \)).
Using Factor Theorem
Since \( (x + 2) \) is a factor, we have:
\[ P(-2) = 0 \]
Substituting \( x = -2 \):
\[ a(-2)^3 + b(-2)^2 + (-2) - 6 = 0 \]
\[ -8a + 4b - 2 - 6 = 0 \]
\[ -8a + 4b - 8 = 0 \]
\[ -8a + 4b = 8 \]
\[ 2b - 4a = 4 \quad \text{(1)} \]
Using Remainder Theorem
Next, substituting \( x = 2 \) for the remainder:
\[ P(2) = 4 \]
Substituting \( x = 2 \):
\[ a(2)^3 + b(2)^2 + (2) - 6 = 4 \]
\[ 8a + 4b + 2 - 6 = 4 \]
\[ 8a + 4b - 4 = 4 \]
\[ 8a + 4b = 8 \]
\[ 2b + 4a = 4 \quad \text{(2)} \]
Solving the System of Equations
Now we have two equations:
1. \( 2b - 4a = 4 \) (1)
2. \( 2b + 4a = 4 \) (2)
Substituting (1) into (2):
From equation (1):
\[ 2b = 4 + 4a \]
Substituting into equation (2):
\[ 4 + 4a + 4a = 4 \]
\[ 8a = 0 \]
\[ a = 0 \]
Using \( a = 0 \) in equation (1):
\[ 2b - 4(0) = 4 \]
\[ 2b = 4 \]
\[ b = 2 \]
Final Values
Thus, the values are:
- \( a = 0 \)
- \( b = 2 \)
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If ax^3+bx^2+x-6 has (x+2) as a factor and leaves a remainder 4 when divided by (x-2),find the value of 'a' and 'b'. ANSWER a=0 b=2?
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