Question 370·Hard·Math

0 / 2000

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

A zero of a polynomial is an input that makes its output $0$ . Graph the given polynomial in Desmos and click every $x$ -intercept, where the curve meets the horizontal axis. Then add only the zeros with positive $x$ -values. The tempting slip is to add every intercept, including the negative ones.

Hints

Hint 1
A zero makes the function output $0$ . On a graph, zeros appear at $x$ -intercepts, where the curve meets the horizontal axis. How could you use the given formula to find those points?
Hint 2
Check the whole graph, not only the first crossing you see. A fourth-degree polynomial can have at most four distinct real zeros, so finding four intercepts tells you that you have them all.
Hint 3
Positive describes a zero’s $x$ -value, not the height of the curve near it. Which intercepts lie to the right of $0$ ? Add only those $x$ -values.

Step-by-step

Approach 1: Graph and add the positive intercepts
Step 1
Turn zeros into an equation
A zero is an input that makes $h(x)=0$ , so set the polynomial’s output to $0$ :
$x^4-2x^3-7x^2+8x+12=0.$
Step 2
Find every zero on the graph
Type $h(x)=x^4-2x^3-7x^2+8x+12$ in Desmos. Click the curve, then click its four $x$ -intercepts. Desmos shows $(-2,0)$ , $(-1,0)$ , $(2,0)$ , and $(3,0)$ . Their $x$ -coordinates are zeros. A fourth-degree polynomial has at most four distinct zeros, so these are all of them. Find every zero before choosing which ones the question asks for.
Step 3
Keep only the positive zeros
Positive means greater than $0$ , so keep $2$ and $3$ . The values $-2$ and $-1$ are zeros too, but they don’t meet that condition.
Step 4
Add the zeros you kept
Type $2+3$ on the next Desmos line. It prints $5$ , so the sum of the positive real zeros is $5$ . Grid in 5.

Approach 2: Reveal the factors by grouping
Step 1
Find a repeated expression
Split $-7x^2$ into $x^2-8x^2$ . The first three terms now make a square, and the next two share $x^2-x$ :
$\begin{aligned}h(x)&=x^4-2x^3+x^2-8x^2+8x+12\\[1.4em]&=(x^2-x)^2-8(x^2-x)+12.\end{aligned}$
That repeated expression makes the fourth-degree polynomial easier to factor.
Step 2
Factor the repeated-expression pattern
Treat $x^2-x$ as one quantity. The numbers $-2$ and $-6$ multiply to $12$ and add to $-8$ , so factor the quadratic pattern:
$h(x)=(x^2-x-2)(x^2-x-6).$
Step 3
Factor both smaller quadratics
For each quadratic, choose two numbers that multiply to its constant and add to $-1$ , the coefficient of $x$ :
$\begin{aligned}x^2-x-2&=(x-2)(x+1),\x^2-x-6&=(x-3)(x+2).\end{aligned}$
So every factor is linear: $h(x)=(x-2)(x+1)(x-3)(x+2)$ .
Step 4
Solve each factor equation
Use the zero-product property: a product is $0$ when at least one factor is $0$ . Set each factor equal to $0$ :
$x-2=0\quad\text{or}\quad x+1=0\quad\text{or}\quad x-3=0\quad\text{or}\quad x+2=0.$
Solve those equations:
$x=2,\ -1,\ 3,\ -2.$
Step 5
Select the positive values
Of those four zeros, only $2$ and $3$ are positive. Leave $-1$ and $-2$ out of the requested sum.
Step 6
Find their sum
Add the two positive zeros:
$2+3=5.$
So the sum of the positive real zeros of $h$ is $5$ . Grid in 5.

Question 370·Hard·Math

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this