Zeros of Polynomials: Chapter Test

Degrees add when polynomials multiply, and an exponent multiplies the degree of what it sits on, so the constant contributes nothing and the exponents contribute $3$, $2$ and $1$.

$\deg p = 3 + 2 + 1 = 6$

Written with nothing collected there are six linear factors, carrying the roots $6, 6, 6, -2, -2, 11$. Only three different numbers appear on that list, so $p$ has three distinct roots, and the multiplicities $3 + 2 + 1$ add back to the degree.

The exponent on $(x + 5)$ is $4$, so $-5$ is a zero of multiplicity $4$. Split the polynomial into that factor and everything else.

$P(x) = (x + 5)^4\,G(x), \qquad G(x) = (x - 2)^3(x + 1), \qquad G(-5) = (-343)(-4) \neq 0$

The multiplicity is even, so $(x + 5)^4$ is positive on both sides of $-5$ and $P$ keeps the sign of $G(-5)$ on both sides. The curve reaches the axis and leaves on the side it came from.

The sum of the roots is $e_1$, and $e_k = (-1)^k \frac{a_{n-k}}{a_n}$ with $k = 1$ gives a single minus sign and a division by the leading coefficient.

$e_1 = -\frac{a_2}{a_3} = -\frac{-4}{5} = \frac{4}{5}$

Both steps are needed: the minus sign comes from the one root each term of that coefficient collects, and the $5$ underneath comes from making the polynomial monic before comparing it with the factorization.

Every coefficient is real, so the conjugate root theorem applies and the conjugate of any root is a root. Conjugation flips the sign of the imaginary part and leaves the real part exactly where it is.

$\overline{-7 + 4i} = -7 - 4i$

The real part $-7$ does not move, because the swap that proves the theorem is $i \mapsto -i$ and nothing else.

A multiplicity counts every copy of the factor, wherever it is hiding, so the quadratic has to be factored before anything is counted.

$p(x) = (x + 5)^2(x + 5)(x - 2) = (x + 5)^3(x - 2)$

The quadratic contributes a third copy, and the leftover factor takes the value $-5 - 2 = -7$ at $-5$, which is not zero, so the count stops at $3$.

The coefficients are rational, so the surd theorem supplies the partner $4 + 3\sqrt{2}$, and the two are distinct, so the quadratic they build divides the polynomial. Their sum is $8$ and their product is a difference of squares.

$\bigl(4 - 3\sqrt{2}\bigr)\bigl(4 + 3\sqrt{2}\bigr) = 16 - 9 \cdot 2 = -2$

A monic quadratic is $x^2$ minus the sum times $x$ plus the product, which is $x^2 - 8x - 2$.

Write out every coefficient first, including the one that is missing: there is no $x$ term, so $a_1 = 0$ rather than absent. With $a_3 = 5$, $a_2 = 20$, $a_1 = 0$,

$e_1 = -\frac{20}{5} = -4, \qquad e_2 = +\frac{0}{5} = 0, \qquad r_1^2 + r_2^2 + r_3^2 = e_1^2 - 2e_2 = 16 - 0 = 16.$

The identity comes from squaring $e_1$, which produces the squares plus twice every pairwise product, so subtracting $2e_2$ leaves the squares alone. A missing term is a constraint, not a gap: it says the pairwise products cancel to exactly zero.

The sign flips at a real zero exactly when that zero has odd multiplicity, and the exponents here are $2$, $1$ and $4$.

$P(x) = \underbrace{(x + 3)^2}_{\text{even}}\ \underbrace{(x - 1)}_{\text{odd}}\ \underbrace{(x - 6)^4}_{\text{even}}$

Only $x = 1$ carries an odd exponent, so the curve crosses there and touches at the other two. A sign change requires a real zero, but a real zero does not force a sign change, and this polynomial has two zeros that decline to flip.

Real coefficients pair a nonreal root with its conjugate at equal multiplicity, so $2 + 5i$ is also a root of multiplicity $2$. That is four of the six places on the root list.

$6 - (2 + 2) = 2$

At most two roots are left, and both could be real, so the ceiling is $2$ and it is reached, for instance by $(x^2 - 4x + 29)^2(x - 1)^2$. Matching the multiplicities is what does the work here: pairing the values alone would have left four places.

Treat it as a quadratic in $x^2$: the two numbers multiplying to $-4$ and adding to $3$ are $4$ and $-1$.

$x^4 + 3x^2 - 4 = (x^2 + 4)(x^2 - 1) = (x - 2i)(x + 2i)(x - 1)(x + 1)$

A factorization is complete only when every factor is linear, and $x^2 + 4$ splits over the complex numbers because $(2i)^2 = -4$. Four linear factors for a degree $4$ polynomial, as the counting demands.

A crossing needs odd multiplicity, so at least $1$; a touch needs even multiplicity, so at least $2$. The real multiplicities therefore total an odd number, and the nonreal roots come in pairs, so the degree has the same parity as that odd total.

$n = \underbrace{(\text{odd}) + (\text{even})}_{\text{real, an odd total}} + \underbrace{2m}_{\text{nonreal}}$

Every possible degree is odd and at least $3$, which rules out $2$, $4$ and $6$. Degree $5$ is reached by $(x + 4)(x - 6)^2(x^2 + 1)$, whose picture is exactly the one described.

The product of all the roots is $e_n$, so use $e_k = (-1)^k \frac{a_{n-k}}{a_n}$ with $n = k = 4$.

$e_4 = (-1)^4\,\frac{a_0}{a_4} = (+1)\,\frac{-22}{2} = -11$

Here $k$ is even, so the factor $(-1)^k$ is $+1$ and the constant term keeps its own sign once it has been divided by $a_4$. A cubic would have put a minus in front of the very same reading, which is why the parity has to be read off each time rather than recalled.

Divide, then divide the quotient, and stop only when a value fails to be zero. Synthetic division of $1, -7, 9, 27, -54$ by $3$ leaves $x^3 - 4x^2 - 3x + 18$; dividing that by $3$ leaves $x^2 - x - 6$; dividing again leaves $x + 2$.

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

That last value is the leftover factor evaluated at $3$, and it is not zero, so the count stops at three. The multiplicities $3 + 1$ add to the degree $4$.

Take the common factor out first, remembering that multiplying by a nonzero constant cannot move a root.

$p(x) = 3i\bigl(x^3 - 2x^2 + 4x - 8\bigr) = 3i\,(x^2 + 4)(x - 2)$

The roots are $2i$, $-2i$ and $2$, so the nonreal ones are a matched pair and one root is real. The theorem still does not apply, because its hypothesis is that every coefficient of $p$ is real and not one of them is. A theorem that does not apply says nothing either way, which is not the same as forbidding the conclusion.

The crossing is the only thing the picture settles at $x = 3$, and a crossing happens at a real zero exactly when the multiplicity there is odd.

$\text{crosses at } r \iff \text{the multiplicity of } r \text{ is odd}$

So the multiplicity is $1$, $3$, $5$ or more, and flatness narrows that only as a suggestion. The degree is not fixed either, since extra multiplicity and extra nonreal pairs are both invisible, and nonreal roots leave no intercept to look for.

The conjugate root theorem does apply, since the coefficients are real, but $2 - \sqrt{6}$ is a real number and conjugation fixes exactly the reals, so it returns the same root and nothing new.

$\overline{2 - \sqrt{6}} = 2 - \sqrt{6}$

The surd partner comes from a different swap, and that swap needs the coefficients to be rational. A real polynomial can have $2 - \sqrt{6}$ as a root without $2 + \sqrt{6}$: for instance $\bigl(x - (2 - \sqrt{6})\bigr)(x - 1)$, whose coefficients are real and irrational.

The degree of $p$ is $3 + 2 = 5$, and $q(2) \neq 0$ means $q$ contributes no further copies of $(x - 2)$, so every root of $q$ is the second distinct root. Since there are exactly two distinct roots in all, $q$ has only one, say $s$, and its two linear factors both carry it.

$p(x) = (x - 2)^3\,a(x - s)^2, \qquad 3 + 2 = 5$

The multiplicities of the distinct roots always add to the degree, so the second root takes the two places the first one did not.

Put the reciprocals over a common denominator. The numerator collects every product of three of the four roots and the denominator is the product of all four.

$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4} = \frac{e_3}{e_4} = \frac{-\frac{4}{2}}{+\frac{-8}{2}} = \frac{-2}{-4} = \frac{1}{2}$

The sign on $e_3$ is negative because $k = 3$ is odd, and the sign on $e_4$ is positive because $k = 4$ is even. The step is legal because $e_4 \neq 0$, which is the same as saying the constant term is not zero, which is the same as saying no root is zero.

A polynomial with this list exists, and writing it down disposes of the option that says otherwise.

$p(x) = a\,\bigl(x - (3 - i)\bigr)^2\bigl(x - (3 + i)\bigr)(x + 2), \qquad a \neq 0$

But real coefficients would force $3 - i$ and $3 + i$ to occur the same number of times, and here one occurs twice while the other occurs once. Multiplying by a nonzero constant cannot change how often a root occurs, so no nonzero constant multiple of $p$ has real coefficients either. A pairing that matches the values but not the multiplicities is not a pairing.

Away from its zeros, $(x + 1)^2$ is positive, and $a$ is negative, so the sign of $P$ is the opposite of the sign of $(x - 4)^m$.

$P(x) = \underbrace{a}_{<\,0}\ \underbrace{(x + 1)^2}_{>\,0}\ (x - 4)^m$

For $x > 4$ the last factor is positive whatever $m$ is. For $x < 4$ it is positive when $m$ is even and negative when $m$ is odd. So $P$ stays negative everywhere except at its zeros exactly when $m$ is even, and when $m$ is odd the curve rises above the axis to the left of $4$. The size of $m$ only flattens the curve near $4$; the parity is what decides the side.