The Factor Theorem: Practice

The factor $x - 1$ vanishes at $x = 1$, so evaluate there.

$P(1) = 1 - 3 + 5 - 3 = 0$

Since $P(1) = 0$, the factor theorem says $x - 1$ is a factor. Testing $x = -1$ would answer a question about the factor $x + 1$ instead.

Evaluate at the value that kills the candidate factor, $x = 2$.

$P(2) = 8 + 2 - 6 = 4$

The remainder on dividing by $x - 2$ is $P(2) = 4$, not $0$, so the division is not exact and $x - 2$ is not a factor. Divisibility of the constant term is never the test; only the evaluation is.

Rewrite the factor in the form $x - c$: since $x + 4 = x - (-4)$, here $c = -4$.

$x + 4 \text{ is a factor of } P(x) \iff P(-4) = 0$

The rule of thumb is to plug in the value that makes the candidate factor vanish, and $x + 4$ vanishes at $x = -4$.

The factor for the zero $c = -3$ is $x - c = x - (-3)$.

$x - (-3) = x + 3$

Check by substitution: $x + 3$ equals $0$ exactly when $x = -3$, matching the given zero. The factor $x - 3$ would correspond to a zero at $+3$.

The theorem converts the root directly into a linear factor.

$P(7) = 0 \quad\Longrightarrow\quad P(x) = (x - 7)\,Q(x)$

So $x - 7$ is a factor. The degree and the constant term are not determined by one root: many polynomials of many degrees vanish at $7$.

A root at $c = -1$ gives the factor $x - c = x - (-1)$.

$P(-1) = 0 \quad\Longrightarrow\quad x + 1 \text{ is a factor}$

Confirm the root: $P(-1) = -1 - 3 + 1 + 3 = 0$. The other options fail the evaluation test: $P(2) = -3$, $P(-3) = -48$, and $P(-2) = -15$, none of which is $0$.

The zero at $0$ contributes the factor $x$ itself, so the product of the three factors is $x(x + 1)(x - 3)$.

$(x + 1)(x - 3) = x^2 - 2x - 3 \quad\Longrightarrow\quad x\left(x^2 - 2x - 3\right) = x^3 - 2x^2 - 3x$

A quick check of one zero: at $x = 3$ the value is $27 - 18 - 9 = 0$, as required.

Each factor vanishes at one point, and a product is zero exactly when some factor is.

$x - 4 = 0,\quad x + 2 = 0,\quad x - 1 = 0 \quad\Longrightarrow\quad x = 4,\ -2,\ 1$

Flip the sign inside each factor to read off its root: $x + 2$ gives $-2$, not $+2$.

The constant factor $3$ is never zero, so it contributes no root; only the linear factors do.

$x + 5 = 0 \text{ or } x - 2 = 0 \quad\Longrightarrow\quad x = -5 \text{ or } x = 2$

A leading coefficient rescales the graph vertically but does not move its $x$-intercepts.

Degrees add when polynomials multiply, so the two factors' degrees must sum to the cubic's degree $3$.

$3 = \deg(x - 2) + \deg Q = 1 + \deg Q \quad\Longrightarrow\quad \deg Q = 2$

Peeling off a linear factor always lowers the degree by exactly one, which is why the leftover quotient of a cubic is a quadratic.

A factor requires the division to come out exact, and here it does not: evaluating at $x = 3$ shows the leftover.

$P(3) = (3 - 3)\,Q(3) + 5 = 5 \neq 0$

Since $P(3) \neq 0$, the factor theorem says $x - 3$ is not a factor. Producing a quotient is automatic for any division; only a zero remainder makes a factor.

The factor theorem is a biconditional, so a nonzero value settles the question in the negative.

$P(4) = 9 \neq 0 \quad\Longrightarrow\quad x - 4 \text{ is not a factor of } P(x)$

For the second option, note $P(x) - 4$ takes the value $5$ at $x = 4$, still nonzero. The expression that would gain the factor is $P(x) - 9$, whose value at $4$ is $0$.