Find whether 2a^3+a^2-5a+2 is divisible completely by 2a+3 or not
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Here, f(x)= x^3 + ax^2- 2x + a + 4
And, g(x)= x + a
We have, x + a = 0
=> x = -a
Now, put the value thus found into f(x) function
f(-a) = (-a)^3 + a(-a)^2 - 2(-a) + a + 4 =0
=> -a^3 + a^3 + 2a + a = -4
=> 3a = -4
=> a = -4/3
And, g(x)= x + a
We have, x + a = 0
=> x = -a
Now, put the value thus found into f(x) function
f(-a) = (-a)^3 + a(-a)^2 - 2(-a) + a + 4 =0
=> -a^3 + a^3 + 2a + a = -4
=> 3a = -4
=> a = -4/3
keshav9284:
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Given,
g(x) = 2a+3
According to factor theorem,
a = (- 3)/2
p(a) = 2a³+a²-5a+2
=>p(-3/2)=2(-3/2)³+(-3/2)²-5(-3/2)+2
=>p(-3/2)=5 ≠0
But,
According to remainder theorem p(a) should be zero.
So 2a + 3 is not completely divisible by p(a).
Hope this will help you.
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g(x) = 2a+3
According to factor theorem,
a = (- 3)/2
p(a) = 2a³+a²-5a+2
=>p(-3/2)=2(-3/2)³+(-3/2)²-5(-3/2)+2
=>p(-3/2)=5 ≠0
But,
According to remainder theorem p(a) should be zero.
So 2a + 3 is not completely divisible by p(a).
Hope this will help you.
If you like my answer
Please mark it as brainliest
And
Be my follower if possible.
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