Graphing Polynomial Functions: Additional Practice (Set 1)

Each division with remainder $0$ pulls out one factor of $(x+3)$, and the first nonzero remainder says the factor has run out. Two clean divisions and then a remainder of $5$ means $(x+3)$ divides $P$ exactly twice.

$P(x) = (x+3)^2 G(x), \qquad G(-3) = 5 \neq 0$

That last remainder is the value of the quotient at $-3$, by the Remainder Theorem, and it is nonzero, so the multiplicity is exactly $2$, not $3$.

A multiplicity of $2$ is even, so $(x+3)^2$ stays positive on both sides of $-3$ and $P$ keeps the sign of $G(-3)$ on both sides. The curve comes down to the axis at $x = -3$ and leaves on the side it arrived from, so it touches and turns back rather than crossing. Repeated synthetic division is exactly how you measure a multiplicity in practice.

A crossing happens at a real zero of odd multiplicity, so check the parity of each exponent.

$x = 1 \ (m = 2, \text{ even}), \qquad x = -2 \ (m = 3, \text{ odd})$

Only $x = -2$ has odd multiplicity, so the graph crosses there and nowhere else. At $x = 1$ it touches the axis and turns back. The crossing at $-2$ is flattened, since the multiplicity is $3$, but a flattened crossing is still a crossing.

The interval $2 < x < 4$ has no zero of $P$ inside it, so a single test point decides the sign of the whole interval. Take $x = 3$.

$P(3) = (4)(1)(-1) = -4 < 0$

Exactly one factor is negative there, so the product is negative and the curve runs below the axis across the entire interval.

Substitute $x = 0$ into every factor, keeping the leading constant.

$P(0) = -2(0-3)^2(0+1) = -2(9)(1) = -18$

The graph passes through $(0, -18)$. Squaring $-3$ makes it positive, so the only minus sign that survives is the one out front.

The Rational Root Theorem offers the divisors of $-6$, since the leading coefficient is $1$. Test $x = 2$.

$P(2) = 8 + 8 - 10 - 6 = 0$

So $(x-2)$ is a factor. Synthetic division by $2$ on the coefficients $1, 2, -5, -6$ gives the depressed polynomial $x^2 + 4x + 3$, which factors easily.

$P(x) = (x-2)(x+1)(x+3)$

The three real zeros $2$, $-1$, and $-3$ are all simple, so the graph crosses the axis at three distinct points.

There are exactly $7$ roots counted with multiplicity, and the real zeros use up $2 + 1 = 3$ of them.

$7 - 3 = 4$

So $4$ roots are non-real, which is two conjugate pairs. The answer had to be even, since non-real roots of a real polynomial always pair off with their conjugates, and that check rules out $3$ and $5$ immediately.

Write $P(x) = (x+2)^2 G(x)$ with $G(x) = x - 1$, and evaluate $G$ at the zero.

$G(-2) = -2 - 1 = -3 < 0$

Near $x = -2$ the factor $G$ keeps that negative sign, and $(x+2)^2$ is positive on both sides of $-2$. A positive times a negative is negative, so $P$ is negative just to the left of $-2$, and by the same reasoning just to the right of it as well. That is the even-multiplicity picture: the curve rises to touch the axis from below and drops back down.

Read the picture. The curve touches the axis at $x = -1$ and at $x = 2$ without ever crossing, so both zeros have even multiplicity. Both arms point downward, so the degree is even and the leading coefficient is negative.

$P(x) = -(x+1)^2(x-2)^2$

The version without the minus sign has both arms rising, which is upside down. The choice with exponents $1$ and $3$ crosses at both zeros rather than touching. The last choice has degree $3$, so its arms would point in opposite directions.

Factor out the common $x^2$ first, then recognize the perfect square that is left.

$x^4 - 4x^3 + 4x^2 = x^2(x^2 - 4x + 4) = x^2(x-2)^2$

Both zeros have multiplicity $2$, which is even, so the graph touches the axis at both and turns back at both. In fact $P(x) = \left[x(x-2)\right]^2 \geq 0$, so the curve never goes below the axis at all.

The factor $x^2 + 4$ is positive for every real $x$, since a square is never negative, so it never affects the sign. The sign of $P$ is the sign of $(x-3)^3$.

$(x-3)^3 > 0 \iff x - 3 > 0 \iff x > 3$

An odd power keeps the sign of its base, so $P$ is positive exactly when $x > 3$ and negative when $x < 3$. The single real zero at $3$ has multiplicity $3$, odd, so the graph crosses there, which is precisely the sign change we just found.

Three of the four come straight off the factored form. The real zeros $-2$, $1$, and $4$ are the x-intercepts. Substituting $x = 0$ gives the y-intercept.

$P(0) = (2)(-1)(-4) = 8$

And the leading term $x \cdot x \cdot x = x^3$ has odd degree with a positive coefficient, which fixes the end behavior: down on the left, up on the right.

The dip is a different matter. Testing $x = 2$ gives $P(2) = (4)(1)(-2) = -8$, so the curve does sink below the axis between $1$ and $4$ and therefore has a lowest point in there, but nothing in this chapter locates it. Finding where the curve stops falling and starts rising means finding where its slope is zero, which is a calculus question.

Two turning points would have been exceptions, and this is neither of them. A parabola's vertex sits halfway between its roots by symmetry, and a real zero of even multiplicity is a turning point at the exactly known coordinates $(r, 0)$. This dip belongs to a cubic and sits off the axis, so it is out of reach. Beware the tempting guess that it bottoms out at the midpoint $x = 2.5$. Arithmetic alone refutes that.

$P(2.5) = (4.5)(1.5)(-1.5) = -10.125, \qquad P(2.75) = (4.75)(1.75)(-1.25) = -10.390625$

The curve is already lower at $2.75$ than at the midpoint, so the midpoint is not the bottom. Evaluation can refute a wrong guess, but it can never hand you the exact one.

Count the roots first as a check: the conjugate pair contributes $2$ and the double root contributes $2$, giving $2 + 2 = 4$, which matches the degree, so the list is complete.

$\text{real zeros}: \ x = 3 \ \text{only}$

The roots $1 \pm i$ are not real, so they are not points on the x-axis and produce no intercept. The graph therefore meets the axis at exactly $1$ point, and since the multiplicity of $3$ is even, it touches there rather than crossing.