If the zeroes of the polynomial f(x) = x3 – 12x2 + 39x + a are in AP, find the value of a.

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Let p-1,p,p+1 be the roots of given polynomial Now p-1+p+p+1=-b/a1 (P-1)p(p+1)=-d/a1 Compare given polynomial with a1x^3+bx^2+cx+d=0 We get a1=1, b=-12, c=39, d=a 3p=-(-12) P=12/3=4 Therefore (4-1), 4, (4+1)= 3, 4, 5 are zeroes of the polynomial (p-1)p(p+1)=-a 3×4×5=-a -60=a
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a = 28
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Let the zeroes be A-D , D and A+D ( as zeroes of the polynomial are in AP) then sum of zeroes = -(coeff. of x2)/(coeff of x3) = 12 A-D+D+A+D = 12, 3A=12 A=4 Since, A is a zero of the polynomial, it satisfy the eqn. p(A)=0 therefore placing x=A in eqn, we get A3-12A2+39A+a = 64-192+156+a = 0 ...
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28
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Polynomial: f(x)=x3−12x2+39x+a Zeroes are in A.P., so assume them as: A−D,D,A+D The sum of zeroes = - ( coefficient of x2 \coefficient of x3) (A−D)+A+(A+D)=-(-12)\1⇒3A=12 ⇒A=4 So, one of the zeroes is 4 . Now , f(4)= (4)3−12⋅(4)2+39⋅4+a=64−192+156+a=28+a Since...
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Polynomial: f(x)=x3−12x2+39x+a Zeroes are in A.P., so assume them as: A−D,D,A+D The sum of zeroes = - ( coefficient of x2 \coefficient of x3) (A−D)+A+(A+D)=-(-12)\1⇒3A=12 ⇒A=4 So, one of the zeroes is 4 . Now , f(4)= (4)3−12⋅(4)2+39⋅4+a=64−192+156+a=28+a Since , 4 is a zeroes of the given polynomial then f (4)=0 ⇒28+a=0 ⇒a=−28 read less
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renu sundriyal A=28
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