The Factor Theorem: Practice

Each factor contributes one equation. From $P(1) = 0$: $1 + a + b + 2 = 0$, so $a + b = -3$. From $P(-2) = 0$: $-8 + 4a - 2b + 2 = 0$, so $4a - 2b = 6$, that is $2a - b = 3$.

$a + b = -3, \qquad 2a - b = 3 \quad\Longrightarrow\quad 3a = 0$

So $a = 0$ and $b = -3$. Indeed $x^3 - 3x + 2 = (x - 1)^2(x + 2)$, which contains both required factors.

Three distinct roots of a cubic use up all its linear factors, so $P(x) = a(x - 1)(x - 2)(x - 3)$ for some constant $a$. The value at $0$ finds it:

$P(0) = a(-1)(-2)(-3) = -6a = 12 \quad\Longrightarrow\quad a = -2$

Then $P(4) = -2\,(3)(2)(1) = -12$.

The two factor conditions are $P(1) = 1 + a + b = 0$ and $P(2) = 16 + 4a + b = 0$. Subtract the first from the second:

$15 + 3a = 0 \quad\Longrightarrow\quad a = -5, \qquad b = -1 - a = 4$

The polynomial is $x^4 - 5x^2 + 4 = \left(x^2 - 1\right)\left(x^2 - 4\right)$, whose zeros are $\pm 1$ and $\pm 2$: the even powers made the negatives come along for free.

The factor theorem works at any degree, so evaluate at $x = -1$ and use $(-1)^{101} = -1$, since $101$ is odd.

$P(-1) = -1 - m + 3 = 2 - m = 0$

So $m = 2$. No division of a degree-$101$ polynomial is needed; one evaluation replaces it.

Evaluate each at $x = -1$, remembering that even powers of $-1$ give $1$ and odd powers give $-1$.

$(-1)^{50} - 1 = 1 - 1 = 0$

So $x^{50} - 1$ has the factor $x + 1$. The others give $(-1)^{51} - 1 = -2$, then $1 + 1 = 2$, then $-1 + 2 = 1$, all nonzero.

Real coefficients force the conjugates $-i$ and $-2i$ to be zeros too, giving four factors that pair into real quadratics.

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

Multiplying, $\left(x^2 + 1\right)\left(x^2 + 4\right) = x^4 + 5x^2 + 4$. Both signs are plus, since each conjugate product is a sum of squares.

Peel off the known zero with synthetic division on $1, -7, 17, -15$ by $c = 3$: the rows give $1$, then $-7 + 3 = -4$, then $17 - 12 = 5$, then $-15 + 15 = 0$, so the quotient is $x^2 - 4x + 5$.

$x = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i$

The nonreal zeros are the conjugate pair $2 + i$ and $2 - i$, exactly as real coefficients require.

Expand the right side: $(x - 2)^2(x - r) = \left(x^2 - 4x + 4\right)(x - r) = x^3 - (r + 4)x^2 + (4r + 4)x - 4r$. The left side has no $x^2$ term, so its coefficient must vanish.

$r + 4 = 0 \quad\Longrightarrow\quad r = -4$

Then $b = -4r = 16$ (and $a = 4r + 4 = -12$). Indeed $x^3 - 12x + 16 = (x - 2)^2(x + 4)$.

Test $x = 1$: $1 - 1 + 4 - 4 = 0$, so $x - 1$ is a factor. Synthetic division on $1, -1, 4, -4$ by $1$ gives $1$, then $-1 + 1 = 0$, then $4 + 0 = 4$, then $-4 + 4 = 0$.

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

The quotient $x^2 + 4$ has discriminant $-16 < 0$, so it has no real roots and cannot be broken into real linear factors: over the reals, this is complete.

From the real factorization $(x - 1)\left(x^2 + 4\right)$, finish the quadratic over the complex numbers: its zeros satisfy $x^2 = -4$.

$x = \pm 2i \quad\Longrightarrow\quad x^2 + 4 = (x - 2i)(x + 2i)$

So the cubic splits into three linear factors, $(x - 1)(x - 2i)(x + 2i)$, with the nonreal pair being conjugates of each other.

Each distinct zero peels off one linear factor, and each peel lowers the quotient's degree by exactly one.

$P(x) = (x - r_1)(x - r_2)\cdots(x - r_k)\,Q(x), \qquad \deg Q = 6 - k \geq 0$

So $k \leq 6$, and the bound is achievable, for example by $(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$. The ceiling is a maximum, not a guarantee: a degree-$6$ polynomial may also have fewer real zeros, or none.

Compute the value being subtracted: $P(2) = 8 + 4 + 1 = 13$. Then the shifted polynomial vanishes at $2$:

$P(2) - 13 = 0$

So $x - 2$ is a factor of $P(x) - 13$ by the factor theorem. In general $x - c$ always divides $P(x) - P(c)$, which is just the master identity $P(x) = (x - c)Q(x) + P(c)$ rearranged.