Graphing Polynomial Functions: Practice

Only real zeros are x-intercepts. The first two factors give the real zeros $1$ and $-1$. For the third factor, ask whether $x^2 + 4 = 0$ has a real solution.

$x^2 = -4 \quad \Longrightarrow \quad \text{no real } x, \text{ since a real square is never negative}$

Its two roots are the non-real pair $2i$ and $-2i$, which never appear on the x-axis. So the graph has exactly $2$ x-intercepts, even though the polynomial has $4$ roots counted with multiplicity.

Substitute $x = 2$ and track the sign of each factor.

$P(2) = (2-1)(2-4) = (1)(-2) = -2$

One factor is positive and one is negative, so the product is negative. At $x = 2$ the curve sits below the x-axis.

A touch happens at a zero of even multiplicity, so look for an even exponent.

$P(x) = (x+1)^1(x-2)^{\,2}(x+5)^1$

Only the factor $(x-2)$ has an even exponent, and it vanishes at $x = 2$. The zeros $-1$ and $-5$ have multiplicity $1$, which is odd, so the graph crosses at both of them. Note that $x = -2$ is not a zero at all.

Set each factor to zero: $x = 3$ from the first, and $x = -1$ from the second.

$\text{distinct zeros}: \ 3 \ \text{ and } \ -1$

That is $2$ different numbers, so the graph has $2$ x-intercepts. Counted with multiplicity there are $3$ roots, since $x = 3$ is counted twice, but the question asks how many are distinct.

Add the exponents, since the leading term is the product of the leading terms.

$x^2 \cdot x^3 = x^{2+3} = x^5$

The degree is $5$. Multiplying the exponents would give $6$, which is the standard trap here.

There is no zero to the right of $3$, so $P$ holds one sign on the whole interval $x > 3$, and a single test point settles it. Take $x = 4$.

$P(4) = (3)(2)(1) = 6 > 0$

All three factors are positive once $x$ passes $3$, so the product is positive there.

A test point has to lie strictly inside the interval, and it must not be a zero of $P$. The zeros here are $-2$, $1$, and $3$.

$P(0) = (2)(-1)(-3) = 6 > 0$

Only $x = 0$ is inside the interval, and it gives a nonzero value, so it reports a genuine sign. Feeding in $-2$ or $1$ would return $0$, which reports no sign at all, and $3$ is outside the interval.

Touching without crossing requires a zero at $5$ of even multiplicity.

$(x-5)^2 \geq 0 \quad \text{for every real } x$

The square never goes negative, so that graph comes down to the axis at $x = 5$ and turns back. The first power and the third power both have odd exponents, so those graphs cross at $5$, and $(x+5)$ does not even have a zero at $5$.

The Fundamental Theorem of Algebra says a polynomial of degree $n$ has exactly $n$ roots in the complex numbers, counted with multiplicity.

$n = 7 \quad \Longrightarrow \quad 7 \text{ roots counted with multiplicity}$

The count never depends on how many happen to be real. What does depend on that is the number of x-intercepts, since only the real roots show up on the graph.

The factor $(x-1)$ gives the real zero $x = 1$. For the quadratic factor, check whether it has a real zero.

$x^2 + 9 = 0 \quad \Longrightarrow \quad x^2 = -9, \ \text{ which no real } x \text{ satisfies}$

Its roots are $3i$ and $-3i$, a non-real pair, and non-real roots produce no x-intercepts. So the graph meets the x-axis at exactly $1$ point.

Substitute $x = 0$. Every term with an $x$ in it dies, leaving only the constant term.

$P(0) = 2(0)^3 - 5(0) + 4 = 4$

So the y-intercept is $(0, 4)$. For a polynomial written in standard form, the y-intercept is always the constant term.

A crossing needs odd multiplicity, and the smallest odd multiplicity is $1$. A touch needs even multiplicity, and the smallest even multiplicity is $2$.

$1 + 2 = 3$

So the degree is at least $3$, and $P(x) = (x+1)(x-2)^2$ shows that $3$ is actually achievable. A degree of $2$ is impossible, since two zeros with those requirements already need three factors.