For integer-coefficient polynomials, it is often helpful to test small integer values like $x=\pm1,\pm2,\pm3$ in the polynomial to see if any give $0$.

Once you find one value of $x$ that makes the polynomial $0$, factor out the corresponding linear factor (like $(x-a)$) using division, then factor the remaining polynomial further.

After you factor completely and list all zeros, make sure you pick only those that are both real and positive, and then add those selected values.

Type y = x^4 - 2x^3 - 7x^2 + 8x + 12 into Desmos so you can see the graph of $y=h(x)$.

Zoom and pan until you can clearly see where the graph crosses the x-axis. Click on each x-intercept to see its $x$-coordinate; these $x$-coordinates are the real zeros of $h$.

From the list of x-intercepts, note only the ones with positive $x$-values. Add those positive $x$-values together (outside Desmos or using the calculator line) to get the required sum.

A zero (or root) of $h$ is a value of $x$ that makes $h(x)=0$. The problem asks for the sum of the positive real zeros, so we need to:

• Solve $h(x)=0$.
• Identify which solutions are real and positive.
• Add those positive solutions together.

We are given

A common way to start factoring a polynomial like this is to test small integer values (possible rational roots): $x=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$.

Try $x=2$:

Compute step by step:

• $2^4=16$
• $2\cdot2^3=16$, so $16-16=0$
• $7\cdot2^2=28$, so $0-28=-28$
• $8\cdot2=16$, so $-28+16=-12$
• Finally, $-12+12=0$

Since $h(2)=0$, $x=2$ is a root, and $(x-2)$ is a factor of $h(x)$.

Divide $h(x)$ by $(x-2)$ (using synthetic or long division) to get the quotient:

Now factor the cubic $x^3-7x-6$ by testing small integer roots again.

Test $x=3$:

so $x=3$ is a root, and $(x-3)$ is a factor. Divide $x^3-7x-6$ by $(x-3)$:

Now factor the quadratic:

Putting all the factors together,

so the zeros are $x=2,3,-1,-2$.

From the factored form, the real zeros are $2,3,-1$, and $-2$.

• The positive real zeros are $2$ and $3$.
• Their sum is

So, the sum of the positive real zeros of $h$ is $5$.