8.2 Solving Polynomial Equations

8.2 Solving Polynomial Equations

Table 11’s second polynomial skill is solving. On AAF that almost always means: write the equation as P(x) = 0, factor completely over the integers, then apply the zero product property. The factoring toolkit is already built in GCF, Grouping, and Special Products and Factoring Cubics and Higher Polynomials. This section is what those factors are for: roots, double roots versus two distinct roots, and the factor/remainder theorems at AAF size — if P(r) = 0, then x − r is a factor.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

Always set the polynomial equal to zero first

The zero product property says: if a product of real numbers is zero, then at least one factor is zero. If A · B = 0, then A = 0 or B = 0 (or both). It does not say anything about A · B = 12. So the first move on a polynomial equation is:

1. Expand if needed so you can see like terms.
2. Move every term to one side: P(x) = 0.
3. Factor P(x) completely.
4. Set each non-constant factor equal to zero.
5. Check each candidate in the original equation.

x(x − 3) = 18 is not solved by x = 18 or x − 3 = 18. Subtract 18 first: x^2 − 3x − 18 = 0, then (x − 6)(x + 3) = 0, so x = 6 or x = −3. Check: 6 · 3 = 18 and (−3)(−6) = 18. Both original products match 18. The illegal zero-product shortcut x = 18 gives 18 · 15 = 270, which is not 18.

Never divide both sides by a variable. x^3 = 4x divided by x becomes x^2 = 4 and drops the root x = 0. Move first: x^3 − 4x = 0 = x(x^2 − 4) = x(x − 2)(x + 2), so x = 0, x = 2, and x = −2. The factor x is the root at zero; dividing by x is how that factor disappears from the solution list.

A constant factor that is never zero can be ignored as an equation. 2(x − 5)(x + 1) = 0 still has only x = 5 and x = −1. The 2 is part of the factorization, not a root.

Double roots versus two distinct roots

A linear factor x − r that appears once gives a simple root x = r. If it appears as (x − r)^2, the equation (x − r)^2 = 0 still has only one solution, x = r, called a double root (multiplicity 2). The graph will bounce there in Graphing Polynomials, Roots, and End Behavior; algebraically, counting solutions without multiplicity still lists r once, but the equation is not the same as (x − r)(x + r) = 0.

That distinction is a classic AAF multiple-choice setup: “which equation has x = 4 as its only solution?”

If the options are unfactored, expand the double-root form: (x − 4)^2 = x^2 − 8x + 16. The opposite-root form is x^2 − 16. Those two quadratics share the root 4 and nothing else about their solution sets. x^2 + 8x + 16 = (x + 4)^2 = 0 has only x = −4, which is the other double-root twin; do not confuse it with “only 4.”

Worked: which equation has x = 4 as its only solution?

Compare x^2 − 16 = 0 with x^2 − 8x + 16 = 0.

x^2 − 16 = (x − 4)(x + 4) = 0 ⇒ x = 4 or x = −4. Two distinct solutions.

x^2 − 8x + 16 = (x − 4)(x − 4) = (x − 4)^2 = 0 ⇒ x = 4 only.

A cubic option such as x^3 − 4x^2 = 0 factors as x^2(x − 4) = 0, so x = 0 (double) and x = 4. Not only 4. Another cubic, (x − 4)^3 = 0, does have only x = 4 (multiplicity 3). If the stem says “only solution” and the options include both a square and a cube of (x − 4), both of those have only x = 4; then you need a further clue such as degree 2, or “double root,” to separate them. On a four-choice CAT, usually only one option has a unique root at 4.

If a stem asks how many distinct real solutions (x − 4)^2(x + 1) = 0 has, the answer is two: 4 and −1. If it asks for the degree, that is 3, and counting with multiplicity also gives 3. Read the question: “only solution,” “number of distinct real solutions,” and “degree” are three different counts.

Remainder and factor theorems (AAF size)

You do not need a full division-algorithm lecture. Two facts are enough for this exam.

Remainder theorem. When a polynomial P(x) is divided by the linear polynomial x − r, the remainder is the number P(r). In particular, you can evaluate P(r) by substitution rather than by long division when the only thing you need is the remainder.

Factor theorem. P(r) = 0 if and only if x − r is a factor of P(x). Equivalently: r is a root of P(x) = 0 if and only if you can write P(x) = (x − r) Q(x) for some polynomial Q of degree one less than P.

The two theorems are the same substitution in different language. Remainder 0 means the linear piece divides evenly, which means it is a factor, which means r is a root.

Worked: is x − 2 a factor of P(x) = x^3 − 5x^2 + 2x + 8?

Compute P(2) = 8 − 20 + 4 + 8 = 0. Yes: remainder 0, so x − 2 is a factor.

If the value had been P(2) = 6, the remainder on division by x − 2 would be 6, and x − 2 would not be a factor. Sign of the tested root matters: P(−2) = −8 − 20 − 4 + 8 = −24 ≠ 0, so x + 2 is not a factor of this cubic even though x − 2 is. Testing r checks the factor x − r, not x + r.

Integer-root search

For a polynomial with integer coefficients, any rational root written in lowest terms p/q has p dividing the constant term and q dividing the leading coefficient. On AAF the polynomial is often monic (leading coefficient 1), so possible rational roots are simply the integer factors of the constant term. Test those integers with the factor theorem until one hits zero, then factor the leftover quadratic with the methods in Factoring Quadratic Trinomials.

Worked cubic: x^3 − 4x^2 − 7x + 10 = 0

Possible integer roots: ±1, ±2, ±5, ±10.

P(1) = 1 − 4 − 7 + 10 = 0. So x − 1 is a factor.

Split off the linear factor by synthetic division (or by undoing distribution):

1 | 1   −4   −7   10
  |      1   −3  −10
    1   −3  −10    0

Leftover quadratic: x^2 − 3x − 10 = (x − 5)(x + 2), because −5 and 2 multiply to −10 and add to −3.

Complete factorization: (x − 1)(x − 5)(x + 2) = 0

Solutions: x = 1, x = 5, x = −2

Check by substituting back into the original cubic: P(5) = 125 − 100 − 35 + 10 = 0 and P(−2) = −8 − 16 + 14 + 10 = 0. All three check. If P(1) had not been 0, you would continue down the short list: P(−1) = −1 − 4 + 7 + 10 = 12 ≠ 0, P(2) = 8 − 16 − 14 + 10 = −12 ≠ 0, and so on. You only need one hit; the leftover quadratic is then ordinary factoring or the quadratic formula from Quadratic Formula and Completing the Square.

Another integer-root cubic: x^3 + 2x^2 − 5x − 6 = 0

Possible roots ±1, ±2, ±3, ±6.

P(2) = 8 + 8 − 10 − 6 = 0. Factor x − 2.

2 | 1   2   −5   −6
  |     2    8    6
    1   4    3    0

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

Solutions: x = 2, x = −1, x = −3

A non-monic leftover is the same idea. After a root r is found, the quadratic ax^2 + bx + c may have a ≠ 1; factor it with the ac method, or use the quadratic formula if it is prime over the integers. Completely factored form over the reals may keep an irreducible quadratic, and then the original equation has fewer real roots than its degree.

Creating, then solving, a polynomial equation

Table 11 includes creating in the same content area as solving. Write the equation from the story, then run the zero-product pipeline.

Worked. A cube plus twice its side length equals 3 times the square of the side, plus 6. Side length s > 0:

s^3 + 2s = 3s^2 + 6

s^3 − 3s^2 + 2s − 6 = 0

Test s = 3: 27 − 27 + 6 − 6 = 0. Factor s − 3.

3 | 1   −3   2   −6
  |      3   0    6
    1    0   2    0

(s − 3)(s^2 + 2) = 0

s = 3, and s^2 + 2 = 0 has no real solutions. The only real (and the only positive) solution is s = 3. The leftover s^2 + 2 is prime over the reals. Reporting three real roots would invent factors that are not there. Completely factored over the reals stops at (s − 3)(s^2 + 2).

Worked (product story). Three consecutive integers have product equal to the middle integer plus 60 — the equation created in Section 8.1 was n^3 − 2n − 60 = 0. Test n = 4: 64 − 8 − 60 = −4 ≠ 0. Test n = 5: 125 − 10 − 60 = 55 ≠ 0. Test n = −4: −64 + 8 − 60 ≠ 0. Possible integer roots of n^3 − 2n − 60 are the factors of 60; n = 4 was close but not a root. You do not need to finish every created cubic on a creating stem. On a solving stem, keep testing the factor list (including n = 5 already ruled out) until P(r) = 0, then split. The point of pairing the two skills is mechanical: write P(x) = 0 before you factor, and do not apply zero product to a product that is still equal to 60.

Solving checklist for AAF

1. Expand if the equation is written as a product equal to a nonzero number, then move everything to one side.
2. Factor out a GCF, including a power of x that is a root at 0.
3. Factor completely: grouping, special products, then integer-root test plus leftover quadratic.
4. Apply the zero product property to each non-constant factor.
5. Distinct roots versus a double root: (x − 4)^2 = 0 has only x = 4; (x − 4)(x + 4) = 0 has two solutions.
6. Use P(r) = 0 as both a check and a way to discover a factor. Remainder P(r) is zero exactly when x − r divides P.
7. Substitute each candidate back into the original equation.

Graphing those roots — bounce versus cross, end behavior from degree and leading coefficient — is the next section. Solving tells you where the graph meets the x-axis; graphing tells you how it meets the axis and what it does at the far left and far right.