Question

Let f(x)=x³+ax²+bx+30. It is divisible by x-5, and when it is divided by x+6, the remainder is -396. Find a and b. Also factorize the polynomial completely.
Solve using factor theorem.
Ans.= a=-6, b=-1; (x+2)(x-3)(x-5)

Answer 1

Since (x-5) is a factor of f(x),
at x=5 f(x)=0
125+25a+5b+30=0
25a+5b=(-155)
5a+b=(-31) -------- eq 1
and
at x=(-6) f(x)=(-396) because of remainder theorem,
-216+36a-6b+30=(-396)
36a-6b=(-396+186)
6a-b=(-35) -------- eq 2
ADDING eq 1 and 2
5a+b+6a-b=(-31-35)
11a=(-66)
a=(-6)
so,
6(-6)-b=(-35)
-b=1
b=(-1)

So,f(x)=x^3-6x^2-x+30
f(x)=(x-5)(x^2-x-6) (by dividing f(x) by x-5)
f(x)=(x-5)(x^2-3x+2x-6)
f(x)=(x-5)(x-3)(x+2)

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