Find the values of p and q so that 1 and −2 are the zeroes of the polynomial x3 + 10 x2 + px + q and then find its third zero.
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f(1)=(1)3+10 (1) 2 +p(1) +q = 0
=3+20+p+q=0
=23 + p + q = 0
p+ q = -23
f(-2)=(-2)3+10(-2)2+p(-2)+q=0
=-6+(-40)-2p+q=0
-46-2p+q=0
-2p+q=46
p+q=46/2
p+q=23
=3+20+p+q=0
=23 + p + q = 0
p+ q = -23
f(-2)=(-2)3+10(-2)2+p(-2)+q=0
=-6+(-40)-2p+q=0
-46-2p+q=0
-2p+q=46
p+q=46/2
p+q=23
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