The polynomial function is defined by What is the sum of the positive real zeros of ?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 93 | Free SAT Prep | Aniko

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Question 93·Hard·Nonlinear Functions

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The polynomial function $h$ is defined by

h(x)=x^4-2x^3-7x^2+8x+12

What is the sum of the positive real zeros of $h$ ?

Your answer

(Express the answer as an integer)

For quartic polynomials with integer coefficients, a fast SAT approach is to look for rational roots using small integer tests (like $\pm1,\pm2,\pm3$ ). Once you find one root, factor out the corresponding linear factor using synthetic or long division, then repeat the process on the reduced polynomial until it’s fully factored. Finally, carefully read the question: if it asks for only positive real zeros or a sum/product subset, filter your solutions accordingly before doing the final arithmetic.

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Connect zeros to the equation

A zero of $h$ is a value of $x$ that makes $h(x)=0$ . Rewrite the problem as solving $x^4-2x^3-7x^2+8x+12=0$ and then focusing only on positive real solutions.

Start by searching for easy roots

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 $0$ .

Use factoring step by step

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.

Finish by selecting and adding

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.

Desmos Guide

Enter the polynomial

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

Identify the x-intercepts

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 $h$ .

Select positive intercepts and compute their sum

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.

Step-by-step Explanation

Clarify what the question is asking

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.

Look for rational roots using simple integer tests

We are given

h(x)=x^4-2x^3-7x^2+8x+12.

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 $x=2$ :

h(2)=2^4-2\cdot2^3-7\cdot2^2+8\cdot2+12.

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 $h(x)$ .

Factor the polynomial completely using division and more root tests

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

\frac{x^4-2x^3-7x^2+8x+12}{x-2}=x^3-7x-6.

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

Test $x=3$ :

3^3-7\cdot3-6=27-21-6=0,

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

\frac{x^3-7x-6}{x-3}=x^2+3x+2.

Now factor the quadratic:

x^2+3x+2=(x+1)(x+2).

Putting all the factors together,

h(x)=(x-2)(x-3)(x+1)(x+2),

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

Select the positive real zeros and find their sum

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

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

2+3=5.

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

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

Hints

Desmos Guide

Step-by-step Explanation