find the value of p and q so that 1, -2 are the zeros of the polynomial f(x) = x3+10x2+px+q and find its third zero
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If f(x) =1
f(1)=1'''+10.1''+p + q
f(1)=1 + 10 + p + q
f(1) =p+ q +11..........eqn.1
When f(-2)
f(-2)=-2''' +10(-2'')+p (-2) +q
f(-2) = - 8 +40 - 2p + q
f(-2) = - 2p + q +32......eqn.2
On subtracting eqn 1 and 2 we hav
i. e, f(1) - f(-2)=p+q+11 +2p - q - 32
=3p - 21
Or 3p =21
P =21/3=7.
Putting, p=7,, in eqn. 1
I. e, p + q = - 11
7 + q =-11
q = - 11 - 7=-18
f(1)=1'''+10.1''+p + q
f(1)=1 + 10 + p + q
f(1) =p+ q +11..........eqn.1
When f(-2)
f(-2)=-2''' +10(-2'')+p (-2) +q
f(-2) = - 8 +40 - 2p + q
f(-2) = - 2p + q +32......eqn.2
On subtracting eqn 1 and 2 we hav
i. e, f(1) - f(-2)=p+q+11 +2p - q - 32
=3p - 21
Or 3p =21
P =21/3=7.
Putting, p=7,, in eqn. 1
I. e, p + q = - 11
7 + q =-11
q = - 11 - 7=-18
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