If 1 and-2 are zeroes of polynomial p(x)= x3+10x2+ax+b then find the value of b
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put x=1
then p(1)=1^3+10*1+a*1+b
0=1+10+a+b
a+b=-11........eq1
now put x=-2
then p(-2)=(-2)^3+10(-2)^2+a*-2+b
0=-8+10*4-2a+b
0=-8+40-2a+b
0=32-2a+b
32=2a-b.........eq2
on solving both equations we get
a=7
by putting this value of a in eq1 we get
b=(-18)
then p(1)=1^3+10*1+a*1+b
0=1+10+a+b
a+b=-11........eq1
now put x=-2
then p(-2)=(-2)^3+10(-2)^2+a*-2+b
0=-8+10*4-2a+b
0=-8+40-2a+b
0=32-2a+b
32=2a-b.........eq2
on solving both equations we get
a=7
by putting this value of a in eq1 we get
b=(-18)
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