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Find zero of p(x)=x^3-12x^2+39x-28 if it is given zeros are a-d , a , a-d
nussu2000:
i think it should be a-d , a , a+d
yes
sorry by mistake i typed it
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p(x) = x^3 - 12 x^2 + 39 x - 28
a0=1 a1= -12 a2= 39 a3= -28
S1 = sum of roots is = a-d+a+a+d= -a1/a0
=> 3a = - (-12)/1
=>3a = 12
=>a = 4
by horner's method (or) synthetic division
a0=1 a1= -12 a2= 39 a3= -28
S1 = sum of roots is = a-d+a+a+d= -a1/a0
=> 3a = - (-12)/1
=>3a = 12
=>a = 4
by horner's method (or) synthetic division
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