A polynomial function #f(x) = ax^3+bx^2-28x+15# is divisible by #(x+3)# and #f(x)+60# is divisible by #(x-3)#. How do you find #a, b# and solve #f(x) = 0# ?

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Answer 1 · 2015-06-23T21:05:49

Derive linear equations for #a# and #b# to find #a=2#, #b=-5# and hence:

#f(x) = (2x-1)(x-5)(x+3)#

giving roots of #f(x) = 0#

#x=1/2#, #x=5# and #x=-3#

Looking at the coefficients of #x^3#, #x^2# and the constant term,
we find:

[1] #ax^3+bx^2-28x+15#

#=(ax^2+(b-3a)x+5)(x+3)#

#=ax^3+bx^2+(3b-9a+5)x+15#

[2] #ax^3+bx^2-28x+15+60#

#=(ax^2+(b+3a)x-25)(x-3)#

#=ax^3+bx^2-(3b+9a+25)x+75#

Then looking at the coefficient of #x# we get:

[1a] #3b-9a+5 = -28#

[2a] #3b+9a+25 = 28#

Add [1a] and [2a] to get:

#6b+30 = 0#

So #b = -5#

Subtract [1a] from [2a] to get:

#18a+20 = 56#

So #a=2#

From [1] we then have:

#f(x) = (2x^2-11x+5)(x+3)#

#=(2x-1)(x-5)(x+3)#

So the roots of #f(x) = 0# are #x=1/2#, #x=5# and #x=-3#