The Factor Theorem: Additional Practice (Set 2)

Apply the factor theorem and watch what happens to $k$.

$P(2) = 8 + 4k - 4k - 8 = 0$

The $k$ terms cancel identically, so $P(2) = 0$ no matter what $k$ is, and $x - 2$ is a factor for every real $k$. The polynomial family is built so that the parameter never disturbs the root at $2$.

Evaluate the factorization at $x = 7$ and solve for the unknown value.

$28 = P(7) = (7 - 3)\,Q(7) = 4\,Q(7)$

So $Q(7) = 7$. The identity holds at every input, not just at the root, so it can be read as an equation about values.

Evaluate the identity at $x = 1$, using $P(1) = 1 - 7 + 6 = 0$.

$0 = P(1) = (1 - 2)\,Q(1) = -Q(1)$

So $Q(1) = 0$. The reasoning is the peeling argument in miniature: a root of $P$ that is not the root already peeled must live inside the quotient. (Indeed $Q(x) = x^2 + 2x - 3 = (x + 3)(x - 1)$.)

Evaluate at $x = -1$, the root of the candidate factor. Since $99$ is odd, $(-1)^{99} = -1$.

$(-1)^{99} + 1 = -1 + 1 = 0$

So $x + 1$ is a factor of $x^{99} + 1$. The same argument shows $x + 1$ divides $x^n + 1$ for every odd $n$.

Here the unknown appears in both the polynomial and the factor, but the test is the same: substitute $x = k$.

$P(k) = k^3 - k \cdot k^2 - 4k + 4 = -4k + 4 = 0$

The cubic terms cancel, leaving $k = 1$. Check: $x^3 - x^2 - 4x + 4$ at $x = 1$ is $1 - 1 - 4 + 4 = 0$.

The zeros give $P(x) = a(x + 3)(x - 2)(x - 5)$, and $P(0)$ finds the scale.

$P(0) = a\,(3)(-2)(-5) = 30a = 60 \quad\Longrightarrow\quad a = 2$

Then $P(1) = 2\,(4)(-1)(-4) = 32$.

Multiplying a polynomial by a nonzero constant changes none of its zeros, since $5 \neq 0$ contributes no root.

$5(x - 1)(x - 2) = 0 \iff x = 1 \text{ or } x = 2$

The others differ: $(x + 1)(x + 2)$ and $x^2 + 3x + 2$ both vanish at $-1$ and $-2$, and the cubic option has the extra zero $3$. This is why a zero list alone never picks out one polynomial: it is only when a polynomial of degree $n$ carries $n$ different zeros, as this one does, that they pin it down, and even then a constant factor stays free.

Two factors give two equations. From $P(1) = 0$: $a + b - 3 + 5 = 0$, so $a + b = -2$. From $P(-1) = 0$: $-a + b + 3 + 5 = 0$, so $-a + b = -8$. Subtract the second equation from the first:

$2a = 6 \quad\Longrightarrow\quad a = 3, \qquad b = -5$

Check both: $3 - 5 - 3 + 5 = 0$ and $-3 - 5 + 3 + 5 = 0$.

Set $P(2) = 0$ and solve for $k$.

$P(2) = 16 - 24 + 2k - 10 = 2k - 18 = 0$

So $k = 9$. The evaluation handles the quartic as easily as a quadratic; degree only changes the arithmetic, not the method.

Each distinct zero peels off a linear factor and drops the degree by one, so a degree-$3$ polynomial cannot absorb four peels.

$P(x) = (x - r_1)(x - r_2)(x - r_3)(x - r_4)\,Q(x) \quad\Longrightarrow\quad \deg P \geq 4$

No constant $c$ can change the degree, so four distinct zeros are impossible. The last option overreaches in the other direction: a cubic can have three real zeros, for example $x^3 - x$.

Three distinct zeros of a cubic force the form $P(x) = a(x + 1)(x - 1)(x - 2)$. Evaluate at $x = 3$:

$P(3) = a\,(4)(2)(1) = 8a = 16$

So $a = 2$. The zeros fixed the shape; the one extra data point fixed the scale.

Nonreal zeros of real-coefficient polynomials arrive in conjugate pairs, because conjugating the equation $P(i) = 0$ leaves the real coefficients fixed.

$P(i) = 0 \quad\Longrightarrow\quad P\left(\overline{i}\right) = P(-i) = 0$

So $-i$ must be a zero. Nothing forces any particular real number to be the third zero; that one is free.