Equations and polynomials? II (easier)

Multiply equation $2$ by $a$ , we get $a^2b+ac=74a$

or, $a^2(15-a)+120=74a$ (using equations $1$ and $3$ )

or, $a^3-15a^2+74a-120=0$

Sum of all possible values of $a$ is $\boxed{15}$ .

1. $a+b=15,$

2. $ab+c=74,$

3. $ac=120.$ From 1. $b=15-a,$ , from 3. $c=\frac{120}{a},$

Putting the values on 2

We get $a(15-a)+\frac{120}{a}=74$

or, $a^3-15a^2+74a-120=0$

or, $(a-5)^3-(a-5)=0$

or, $(a-5)[(a-5)^2-1]=0$

or, $(a-5)(a-4)(a-6)=0$

Either $a=5$ ,or $a=4$ ,or $a=6$

Sum of the possible values of $a=4+5+6=15$

We can see that $a$ are roots of the polynomial $P(x)=x^3-15x^2+74x-120.$ Let's try to find rational roots. If $x=\frac{p}{q}$ , then $q^3P(x) \equiv -120q^3 \mod p$ ; since $p,q$ must be coprime, it follows that $p|120$ ; with $q$ as the modulo, we can find that $q=1$ . Trying out the first few divisors of $120$ , we can see that $4,5,6$ are roots and $P(x)=(x-4)(x-5)(x-6).$ Therefore, $a=4,5,6$ are all solutions.

Notice that

$(x+a)(x^2 + bx + c) = x^3 + (a+b)x^2 + (ab+c)x + ac$

We substitute the above values in to get

$x^3 + 15x^2 + 74x + 120 = (x+a)(x^2 + bx + c)(x+a)$

Since $x^3 + 15x^2 + 74x + 120 = (x+4)(x+5)(x+6)$ , we can have the value of $a$ to be one of $4, 5, 6$ and the other quadratic would be the product of he other two brackets. Thus, the sum of all values of $a$ is 15.