Question 342·Medium·Math

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The polynomial function $p$ is defined by

p(x)=x^4 - 5x^3 + 6x^2.

Which of the following points in the $xy$ -plane is NOT an $x$ -intercept of the graph of $p$ ?

An $x$ -intercept is a point where a graph meets the horizontal axis, so the function’s output is $0$ . When a polynomial is given as an equation rather than a graph, graph it in Desmos and click its intercepts. Then evaluate any candidate input you need to check. Watch for a curve that touches the axis without crossing it: that point still counts as an intercept.

Hints

Hint 1
A point $(a,0)$ is an $x$ -intercept only if $p(a)=0$ . The second coordinate is the output to check; it doesn’t mean you should use $x=0$ for every point.
Hint 2
Click the graph’s marked points on the horizontal axis. If the curve touches the axis and turns around, the touching point still has an output of $0$ .
Hint 3
Evaluate the function at a candidate input that isn’t among the intercepts you found. A nonzero output means the graph’s point at that input is not on the horizontal axis.

Step-by-step

Approach 1: Graph the intercepts, then check the remaining input
Step 1
Turn an intercept into a function test
An $x$ -intercept is a point on both the graph and the horizontal axis. For $(a,0)$ to be on the graph, $p(a)$ must equal $0$ . So look for inputs that make the output zero.
Step 2
Read the intercepts from the graph
Type $p(x)=x^4-5x^3+6x^2$ in Desmos. Click the curve, then click its marked points on the $x$ -axis. Desmos shows $(0,0)$ , $(2,0)$ , and $(3,0)$ . The curve touches the axis at $(0,0)$ ; touching still counts.
Step 3
Check the point that is not an intercept
Type $p(5)$ . Desmos prints $150$ , so the graph’s point at input $5$ is $(5,150)$ , not $(5,0)$ . So $(5,0)$ is not an $x$ -intercept. Choice D.

Approach 2: Factor to account for every intercept
Step 1
Pull out the common factor
Each term of $p(x)$ contains $x^2$ , so pull out that common factor, the part shared by every term:
$p(x)=x^2(x^2-5x+6).$
Step 2
Factor the remaining quadratic
The numbers $-2$ and $-3$ multiply to $6$ and add to $-5$ , so the quadratic factors as:
$p(x)=x^2(x-2)(x-3).$
Now every factor that can make the output zero is visible.
Step 3
Find every input with output zero
A product is zero when at least one factor is zero. Set each factor equal to zero:
$x^2=0\quad\text{or}\quad x-2=0\quad\text{or}\quad x-3=0.$
These give $x=0$ , $x=2$ , and $x=3$ . Don’t cancel $x^2$ , or you’ll lose the intercept at $x=0$ .
Step 4
Pick the point outside that list
At $x=5$ , none of the factors is zero, so $p(5)\ne0$ . The point $(5,0)$ is not an $x$ -intercept of $p$ . Choice D.

Question 342·Medium·Math

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