The Factor Theorem: Practice

Synthetic division by $c = 1$ on the row $1, -6, 11, -6$ brings down $1$, then computes $-6 + 1 = -5$, then $11 + (-5) = 6$, then $-6 + 6 = 0$.

$x^3 - 6x^2 + 11x - 6 = (x - 1)\left(x^2 - 5x + 6\right)$

The final $0$ confirms that $1$ really is a zero, and the other entries are the quotient's coefficients.

Write the missing power explicitly: the synthetic row is $1, 0, -13, 12$. Dividing by $c = 1$ gives $1$, then $0 + 1 = 1$, then $-13 + 1 = -12$, then $12 + (-12) = 0$, so the quotient is $x^2 + x - 12$.

$x^2 + x - 12 = (x + 4)(x - 3)$

The complete factorization is $(x - 1)(x + 4)(x - 3)$, with zeros $1$, $-4$, and $3$.

The factor $x + 1$ vanishes at $x = -1$, so require $P(-1) = 0$.

$P(-1) = -2 + k + 5 + 2 = k + 5 = 0$

So $k = -5$. Watch the signs when substituting $-1$: the odd powers flip sign, so $2(-1)^3 = -2$ and $-5(-1) = +5$.

Multiply the three factors $(x + 2)(x - 1)(x - 4)$ in stages. First $(x + 2)(x - 1) = x^2 + x - 2$, then

$\left(x^2 + x - 2\right)(x - 4) = x^3 - 4x^2 + x^2 - 4x - 2x + 8 = x^3 - 3x^2 - 6x + 8$

A check at $x = 1$ gives $1 - 3 - 6 + 8 = 0$, confirming one of the prescribed zeros.

The zeros force the shape $P(x) = a(x + 1)(x - 2)(x - 5)$, and the point pins down $a$. Substitute $x = 0$:

$P(0) = a(1)(-2)(-5) = 10a = 30$

So $a = 3$. Roots alone determine a polynomial only up to this constant; the extra point is what selects it.

Evaluate the polynomial at each candidate's root. At $x = 2$:

$P(2) = 8 - 8 - 10 + 6 = -4 \neq 0$

So $x - 2$ is not a factor. The other three all pass the test: $P(1) = 0$, $P(-2) = 0$, and $P(3) = 0$, which matches the factorization $(x - 1)(x + 2)(x - 3)$.

Pair each zero with its negative, since each pair multiplies into a difference of squares.

$(x - 1)(x + 1) = x^2 - 1, \qquad (x - 2)(x + 2) = x^2 - 4$

Then $\left(x^2 - 1\right)\left(x^2 - 4\right) = x^4 - 5x^2 + 4$. Checking $x = 2$: $16 - 20 + 4 = 0$, as required.

Set the value at $x = -3$ equal to zero.

$P(-3) = -27 + 12 + k = k - 15 = 0$

So $k = 15$. The cube of $-3$ is $-27$, and $-4(-3) = +12$; sign slips on those two terms are the usual source of wrong answers here.

Real coefficients force the conjugate $-3i$ to be the other zero, so the factors are $x - 3i$ and $x + 3i$.

$(x - 3i)(x + 3i) = x^2 - (3i)^2 = x^2 + 9$

The conjugate pair multiplies into a real quadratic with no real roots; $x^2 - 9$ has zeros $\pm 3$, which are real and different.

Real coefficients force $-i$ to join $i$, and the conjugate pair multiplies into a real quadratic.

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

Attaching the real root's factor gives $(x - 2)\left(x^2 + 1\right) = x^3 - 2x^2 + x - 2$. Check at $x = 2$: $8 - 8 + 2 - 2 = 0$.

Evaluate at $x = \tfrac{1}{2}$, the root of the candidate factor.

$P\left(\tfrac{1}{2}\right) = 2 \cdot \tfrac{1}{8} + 3 \cdot \tfrac{1}{4} - 4 + 3 = \tfrac{1}{4} + \tfrac{3}{4} - 1 = 0$

So $2x - 1$ is a factor; in fact $P(x) = (2x - 1)(x + 3)(x - 1)$. Evaluating at $x = 2$ tests the different factor $x - 2$ and gives $15$.

Convert each zero to its factor by flipping the sign: zeros $1$, $4$, $-3$ give factors $x - 1$, $x - 4$, $x + 3$.

$x^3 - 2x^2 - 11x + 12 = (x - 1)(x - 4)(x + 3)$

A quick multiplication check of the constant term: $(-1)(-4)(3) = 12$, matching the polynomial.