Polynomial Division and Roots: Chapter Test

Order the terms by descending power first; nothing may be read off until that is done.

$p(x) = x^5 - 9x^3 - 4x^2 + 7$

The highest power present is $x^5$, and the number sitting on it is $1$, so the leading coefficient is $1$.

The remainder on dividing by $x - 1$ is the value of the polynomial at the input that makes $x - 1$ zero, namely $1$.

$P(1) = 1^3 + 6(1)^2 - 4 = 1 + 6 - 4 = 3$

The missing $x$ term contributes nothing, and the constant term $-4$ is included with its sign.

In $P = DQ + R$ the product $DQ$ must match the dividend's leading term, so the degrees subtract, and the loop runs until the leftover's degree drops below the divisor's.

$\deg Q = 9 - 4 = 5, \qquad \deg R \le 3$

A remainder of degree $4$ would still contain a whole copy of the divisor, so the division would not be finished.

In lowest terms, a rational root $\tfrac{p}{q}$ needs $p$ dividing the constant term $14$ and $q$ dividing the leading coefficient $6$.

$p \in \{1, 2, 7, 14\}, \qquad q \in \{1, 2, 3, 6\}$

The number $\tfrac{4}{3}$ has numerator $4$, and $4$ does not divide $14$, so it cannot be a root. The other three pass both tests, which makes them candidates and nothing more.

Run the tableau with corner $5$ on the coefficients $1, -2, -19, 20$: bring down $1$, then $1 \times 5 = 5$ and $-2 + 5 = 3$, then $3 \times 5 = 15$ and $-19 + 15 = -4$, then $-4 \times 5 = -20$ and $20 - 20 = 0$.

$x^3 - 2x^2 - 19x + 20 = (x - 5)\left(x^2 + 3x - 4\right)$

The last cell is $0$, which confirms the given factor, and the entries before it spell the quotient.

A polynomial graph is unbroken, so it cannot pass from below the axis to above it without touching it.

$f(2) = -9 < 0 < 3 = f(3)$

Somewhere strictly between $2$ and $3$ the value is $0$. The argument locates a zero inside the interval; it neither names it nor counts how many zeros $f$ has elsewhere.

The divisor $x + 2$ is $x - (-2)$, so the remainder is the value at $-2$. Take each power with the input in brackets.

$P(-2) = 2(16) - (-8) - 9(-2) + 6 = 32 + 8 + 18 + 6 = 64$

Run as a tableau the top row is $2,\; -1,\; 0,\; -9,\; 6$, the $0$ holding the missing $x^2$ column open.

A factor $x + 3$ means $P(-3) = 0$, so set the evaluation at $-3$ equal to zero.

$P(-3) = 81 - 27k + 18 + 9 = 108 - 27k = 0 \quad\Longrightarrow\quad k = 4$

Whatever $k$ turns out to be, the highest power present is $x^4$ with coefficient $1$: an even degree with a positive leading coefficient, so both far ends rise.

Feed any positive input into the four terms. Each of $2x^3$, $11x^2$, $17x$ and $6$ is then positive, so their total cannot be zero.

$x > 0 \quad\Longrightarrow\quad 2x^3 + 11x^2 + 17x + 6 > 0$

Half the list is discarded at a stroke, and the search narrows to the negative candidates, where all three roots turn out to sit: $P(x) = (2x + 1)(x + 2)(x + 3)$.

Write the dividend as $x^4 + 2x^3 + 0x^2 - 7x + 5$ and run the loop. Subtracting $x^2\left(x^2 - 3\right)$ leaves $2x^3 + 3x^2 - 7x + 5$; subtracting $2x\left(x^2 - 3\right)$ leaves $3x^2 - x + 5$; subtracting $3\left(x^2 - 3\right)$ leaves $-x + 14$, whose degree is below $2$, so the loop stops.

$x^4 + 2x^3 - 7x + 5 = \left(x^2 - 3\right)\left(x^2 + 2x + 3\right) + (-x + 14)$

Expanding the right side returns the dividend, which is the check worth making every time.

Three roots fix the polynomial only up to a scalar, so write $P(x) = a(x + 5)(x - 1)(x - 4)$ and let the given point choose $a$.

$P(3) = a(8)(2)(-1) = -16a = -32 \quad\Longrightarrow\quad a = 2$

So $P(x) = 2(x + 5)(x - 1)(x - 4)$, and substituting $x = 3$ returns $-32$ as a check.

The sign change locates a real root, and the failed list decides what kind of number it is.

$P(1) = -3 < 0 < 5 = P(2)$

So a real root sits strictly between $1$ and $2$. Since $P$ is monic with constant term $-5$, any rational root would be an integer dividing $5$, and all four of $\pm 1$ and $\pm 5$ have been tested and failed. A real root that cannot be rational is irrational.

Since $3x - 1 = 3\left(x - \tfrac{1}{3}\right)$, run the tableau with corner $\tfrac{1}{3}$ on $6,\; 7,\; -18,\; 9$. The bottom row is $6,\; 9,\; -15,\; 4$, so dividing by the monic factor gives quotient $6x^2 + 9x - 15$ and remainder $4$. Now hand the leftover $3$ to the quotient.

$6x^3 + 7x^2 - 18x + 9 = (3x - 1)\left(2x^2 + 3x - 5\right) + 4$

Only the product term absorbs the $3$, so the remainder stays exactly where it was.

The remainder on dividing by $x + 2$ is the value at $-2$, so set that value equal to $4$.

$16 - 4c - 6 + 10 = 4 \quad\Longrightarrow\quad 20 - 4c = 4 \quad\Longrightarrow\quad c = 4$

Checking, $x^4 - 4x^2 + 3x + 10$ at $-2$ gives $16 - 16 - 6 + 10 = 4$.

Degrees add across a product and the leading terms multiply, so neither expansion nor the lower terms are needed.

$\deg f = 6 + 2 + 1 = 9, \qquad \left(-3x^2\right)^3\left(x^2\right)(-x) = \left(-27x^6\right)\left(x^2\right)(-x) = 27x^9$

The cube of $-3$ is $-27$, and the second minus sign arrives with $-x$, so the two negatives leave a positive leading coefficient.

Divide by $x^2 - 1$, writing the divisor as $x^2 + 0x - 1$ so the columns line up. The three passes give quotient terms $x^2$, $-3x$ and $-10$, and the leftover is $0$.

$x^4 - 3x^3 - 11x^2 + 3x + 10 = \left(x^2 - 1\right)\left(x^2 - 3x - 10\right)$

A remainder of $0$ certifies both parts as factors. Splitting each one, $x^2 - 1 = (x - 1)(x + 1)$ and $x^2 - 3x - 10 = (x - 5)(x + 2)$, so $P(x) = (x - 1)(x + 1)(x - 5)(x + 2)$.

The permitted numerators divide $6$ and the permitted denominators divide $4$, so the list is $\pm 1, \pm 2, \pm 3, \pm 6, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}, \pm\tfrac{1}{4}, \pm\tfrac{3}{4}$. Testing $-\tfrac{1}{4}$:

$P\!\left(-\tfrac{1}{4}\right) = 4\left(-\tfrac{1}{64}\right) - 3\left(\tfrac{1}{16}\right) + \tfrac{25}{4} - 6 = -\tfrac{1}{16} - \tfrac{3}{16} + \tfrac{25}{4} - 6 = 0$

So $-\tfrac{1}{4}$ is a root, and $4x + 1$ is a factor. The other roots are $3$ and $-2$.

Take the two halves in turn. For the first, look for a neighboring pair in the table whose sign changes.

$h(2) = -2 < 0 < 31 = h(3)$

The graph is unbroken, so between those two inputs it has to meet the axis, and that interval is forced.

The pair at $3$ and $4$ is a different matter. Both $31$ and $98$ sit above the axis, and matching signs settle nothing in either direction: a curve may stay clear of the axis across such an interval, and it may equally dip below and come back, meeting the axis twice. So that pair leaves the question open rather than closing it, and the same holds for $h(0)$ with $h(1)$, and for $h(1)$ with $h(2)$.

Each stated remainder is one evaluation, so the two conditions become two linear equations.

$P(2) = 4a + 2b - 4 = 0, \qquad P(-1) = a - b - 13 = -30$

The first tidies to $2a + b = 2$ and the second to $a - b = -17$. Adding them gives $3a = -15$, so $a = -5$ and then $b = 2 - 2(-5) = 12$. Checking, $x^3 - 5x^2 + 12x - 12$ gives $P(2) = 0$ and $P(-1) = -30$.

Shift the whole picture down by $7$ so that the three given inputs become roots. The polynomial $P(x) - 7$ is still a monic cubic, and it vanishes at $-4$, $1$ and $6$, so those three linear factors use up all of its degree.

$P(x) - 7 = (x + 4)(x - 1)(x - 6)$

Now evaluate at $2$: the factors give $(6)(1)(-4) = -24$, so $P(2) = -24 + 7 = -17$.