The Factor Theorem: Additional Practice (Set 1)

Confirm the zero: $P(2) = 8 + 12 - 8 - 12 = 0$. Synthetic division on $1, 3, -4, -12$ by $c = 2$ gives $1$, then $3 + 2 = 5$, then $-4 + 10 = 6$, then $-12 + 12 = 0$, so the quotient is $x^2 + 5x + 6$.

$x^2 + 5x + 6 = (x + 2)(x + 3)$

The complete factorization is $(x - 2)(x + 2)(x + 3)$, with zeros $2$, $-2$, and $-3$.

Require the polynomial to vanish at $x = -2$, keeping careful track of the signs on odd powers.

$P(-2) = 16 - 8k - 12 - 10 + 14 = 8 - 8k = 0$

So $k = 1$. The term $kx^3$ contributes $k(-2)^3 = -8k$, and $5x$ contributes $-10$.

Fold the leading coefficient into the fractional zero's factor: $2\left(x - \tfrac{1}{2}\right) = 2x - 1$, so the polynomial is $(2x - 1)(x + 1)(x - 4)$. Multiply the last two factors first:

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

Then $(2x - 1)\left(x^2 - 3x - 4\right) = 2x^3 - 6x^2 - 8x - x^2 + 3x + 4 = 2x^3 - 7x^2 - 5x + 4$.

Multiply the conjugate pair first, since it produces a real quadratic:

$\bigl(x - (1 + i)\bigr)\bigl(x - (1 - i)\bigr) = (x - 1)^2 - i^2 = x^2 - 2x + 2$

The real zeros give $(x - 1)(x + 1) = x^2 - 1$. Then $\left(x^2 - 2x + 2\right)\left(x^2 - 1\right) = x^4 - 2x^3 + 2x^2 - x^2 + 2x - 2 = x^4 - 2x^3 + x^2 + 2x - 2$.

Set the evaluation at $x = 3$ equal to zero.

$P(3) = 27 - 45 + 3m + 3 = 3m - 15 = 0$

So $m = 5$. Check: $x^3 - 5x^2 + 5x + 3$ at $x = 3$ gives $27 - 45 + 15 + 3 = 0$.

Substitute the value that makes the candidate factor vanish. Solve $3x + 2 = 0$:

$x = -\tfrac{2}{3}$

Since $3x + 2 = 3\left(x + \tfrac{2}{3}\right)$ and the constant $3$ never affects exactness of division, $3x + 2$ is a factor exactly when $P\left(-\tfrac{2}{3}\right) = 0$.

Peel off one known zero: dividing by $x + 1$ via the row $1, -6, 3, 10$ with $c = -1$ gives $1$, then $-6 - 1 = -7$, then $3 + 7 = 10$, then $10 - 10 = 0$, so the quotient is $x^2 - 7x + 10$.

$x^2 - 7x + 10 = (x - 2)(x - 5)$

The quotient's zeros are $2$ and $5$; the new one is $5$, so the three zeros are $-1$, $2$, and $5$.

The conjugate $3 - 2i$ must also be a zero, and its pair with $3 + 2i$ multiplies into a real quadratic:

$\bigl(x - (3 + 2i)\bigr)\bigl(x - (3 - 2i)\bigr) = (x - 3)^2 - (2i)^2 = x^2 - 6x + 13$

Then $(x + 1)\left(x^2 - 6x + 13\right) = x^3 - 6x^2 + 13x + x^2 - 6x + 13 = x^3 - 5x^2 + 7x + 13$.

The two known zeros peel off $(x - 1)(x + 1) = x^2 - 1$. Long division of the quartic by $x^2 - 1$ proceeds in three steps, subtracting $x^2\left(x^2 - 1\right)$, then $-5x\left(x^2 - 1\right)$, then $6\left(x^2 - 1\right)$, and leaves remainder $0$ with quotient

$x^2 - 5x + 6 = (x - 2)(x - 3)$

So the remaining zeros are $2$ and $3$, and the quartet of zeros is $1$, $-1$, $2$, $3$.

Compute the value at the candidate root: $P(1) = 1 - 3 + 7 = 5$. Subtracting exactly that value creates a zero there:

$P(1) - 5 = 0$

So $x - 1$ divides $P(x) - 5$. This is the general pattern: $x - c$ always divides $P(x) - P(c)$, whatever $P$ and $c$ are.

By the factor theorem, $x - c$ is a factor exactly when $c$ is a root, so count the roots.

$x^2 - 7x + 12 = (x - 3)(x - 4)$

The roots are $3$ and $4$, so exactly $2$ values of $c$ work. The ceiling theorem guarantees a quadratic can never do better than two.

One evaluation at $x = 1$ answers the question, no matter how large the degree.

$P(1) = 1^{100} + 1^{50} - 2 = 1 + 1 - 2 = 0$

So $x - 1$ is a factor. The degree being even is irrelevant, and the last option fails for almost every polynomial, for example $x^2$ at $x = 1$ gives $1 \neq 0$.