LCM of functions

Since the lowest common multiple are three factors and has a degree of $3$ , while $f(x)$ , $g(x)$ , and $h(x)$ each has a degree of $2$ . We can assume that each of the three function is composed of two factors. By observation:

$\begin{aligned} f(x) & = (x+8)(x-2) = x^2 + 6x - 16 & \implies a = - 16 \\ g(x) & = (x+6)(x-2) = x^2 + 4x - 12 & \implies b = -12 \\ h(x) & = (x+8)(x+6) = x^2 + 14x + 48 & \implies c = 48 \end{aligned}$

Then the lowest common multiple of $f(x)$ , $g(x)$ , and $h(x)$ is $(x+8)(x-2)(x+6)$ ; and $a+b+c = -16-12+48 = \boxed{20}$ .

The three quadratic functions correspond to the three ways of choosing two factors from the cubic; so for example $(x+8)(x-2)=x^2+6x-16$ and we find $a=-16$ .

Similarly, $b=-12$ and $c=48$ so $a+b+c=\boxed{20}$ .

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Alternatively, note that we must have $f(x)\cdot g(x)\cdot h(x)=\left((x+8)(x-2)(x+6)\right)^2$

The coefficient of $x^4$ on the left-hand side is $a+b+c+164$ ; so we just need to find that coefficient on the right.

Since $(x+8)(x-2)(x+6)=x^3+12x^2+20x-96$ , this coefficient will be $12^2+2\cdot 20=184$ ; so again $a+b+c=20$ .

Here's what we know:

$\begin{cases} f(x)=(x+a_1)(x+a_2) & \quad \text{, where } a_1 + a_2 = 6 \\ g(x)=(x+b_1)(x+b_2) & \quad \text{, where } b_1 + b_2 = 4 \\ h(x)=(x+c_1)(x+c_2) & \quad \text{, where } c_1 + c_2 = 14 \end{cases}$

The common multiples are $(x \color{#3D99F6} +8 \color{#333333} ),(x\color{#D61F06} -2 \color{#333333}),(x\color{#20A900} +6 \color{#333333})$

$\large{ \text{STEP 1}}$ : Find which combinations of the above multiples match each function.

$\large {f(x)}:$ $(\color{#3D99F6} +8 \color{#333333}) + (\color{#D61F06} -2 \color{#333333}) = 6$ .

$\rightarrow \small{\mathbf{a_1}=8 \text{ and } \mathbf{a_2}=-2}$

$\large {g(x)}:$ $(\color{#D61F06} -2 \color{#333333}) + (\color{#20A900} +6 \color{#333333}) = 4$

$\rightarrow \small{\mathbf{b_1}=-2 \text{ and } \mathbf{b_2}=6}$

$\large {h(x)}:$ $(\color{#3D99F6} +8 \color{#333333}) + (\color{#20A900} +6 \color{#333333}) = 14$

$\rightarrow \small{\mathbf{c_1}=8 \text{ and } \mathbf{c_2}=6}$

$\large{ \text{STEP 2}}$ : Find $a, b,$ and $c$ by multiplying the factors of each function.

$\large {f(x)}:$ $(8)(-2)=-16 \Rightarrow \boxed{a=-16}$

$\large {g(x)}:$ $(6)(-2)=-12 \Rightarrow \boxed{b=-12}$

$\large {h(x)}:$ $(8)(6)=48 \Rightarrow \boxed{c=48}$

$\large{ \text{STEP 3}}$ : Find $a+b+c$ .

$(-16)+(-12)+(48)=\boxed{\mathbf{20}}$

When divided by $f(x), g(x)$ and $h(x)$ , the remainder LCM gives is zero. The three functions are of the form $x²+mx+h$ .

$\begin{array}{cc} x²+mx+h)&x³+12x²+20x-96&(x+(12-m)\rightarrow \text{Quotient}\\&- x³+mx+hx& \\ \hline \\&(12-m)x²+(20-h)x-96\\ &-(12-m)x²+(12-m)mx+(12-m)h\\ \hline \\ &((20-h)-(12m-m²))x-(96+(12-m)h)&\rightarrow\text{Remainder}\end{array}$

For $f(x)= x²+6x+a$ , the quotient $x+(12-m)$ is $x+6$ , for $g(x)= x²+4x+b$ , it is $x+8$ , for $h(x)= x²+14x+c$ , it is $x-2$ .

Thus, if the LCM $L(x)$ is divided by $x+6$ , we'd find $f(x)$ , dividing $L(x)$ by $x+8$ we'd have $g(x)$ , dividing $L(x)$ by $x-2$ we'd have $h(x)$ .

This way $f(x)= (x+8)(x-2)\rightarrow a= -16$ $g(x)= (x-2)(x+6)\rightarrow b=-12$ $h(x)= (x+8)(x+6)\rightarrow c=48$

Hence $a+b+c= -16-12+48=\boxed{20}$ .