Polynomial Division

Ever wondered how to break down complicated polynomials? Long division helps you split them into simpler parts.

To divide polynomials using long division, set up the problem similar to regular division. Divide the first term of the dividend by the first term of the divisor, then multiply, subtract, and bring down terms until complete. For example, when dividing $(8x^3 + 34x^2 + x - 70)$ by $(2x + 7)$, you get a quotient of $4x^2 + 3x - 10$.

Synthetic division offers a shortcut when dividing by $(x - k)$. This method eliminates variables and uses just the coefficients. After completing synthetic division, you can express functions in the form $f(x) = (x - k)(q(x)) + r$, where $q(x)$ is the quotient polynomial and $r$ is the remainder.

Quick Tip: In synthetic division, only write the coefficients and drop the variables to make the process faster. This technique saves time on tests!

When using synthetic division, remember to include zero coefficients for any missing terms in your polynomial. This ensures all terms are accounted for during the division process.

Finding Function Values and Verifying Zeros

Synthetic division isn't just for dividing polynomials—it's also a quick way to evaluate functions!

To find function values like $g(-3)$ or $g(3)$, use synthetic division with the value you're evaluating. The remainder at the end of your synthetic division work is the function value. This method is much faster than direct substitution, especially for higher-degree polynomials.

When a number is a zero of a polynomial, the remainder after synthetic division will be zero. For example, if you divide $f(x) = x^3 - 6x^2 - 11x + 30$ by $(x + 2)$, getting a remainder of zero confirms that $-2$ is a zero of the function.

After confirming a zero, you can factor the polynomial further. The quotient you get from synthetic division becomes one factor, and $(x-k)$ is the other. Sometimes you can factor the quotient further, helping you find all zeros of the function.

Remember: When synthetic division gives you a remainder of zero, the divisor $(x-k)$ is a factor of your polynomial, and $k$ is a zero of the function.

Finding Exact Zeros and Complete Factorization

Finding all zeros of a polynomial helps you understand its behavior and graph.

When you know one zero of a polynomial, you can use synthetic division to factor out $(x-k)$, where $k$ is the known zero. After division, you're left with a lower-degree polynomial that's easier to solve. For the function $g(x) = x^3 + 4x^2 - 3x - 12$, if $-4$ is a zero, synthetic division confirms this and gives you $(x+4)(x^2-3)$.

You can further factor this result into $(x+4)(x+\sqrt{3})(x-\sqrt{3})$, giving you all three zeros: $x=-4, -\sqrt{3}, \sqrt{3}$. Each factor corresponds to where the graph crosses the x-axis.

Complex numbers introduce a new dimension to algebra problems. They take the form $a + bi$, where $i = \sqrt{-1}$. To find values like $a$ and $b$ in an equation such as $(a-1)+(b+3)i = 7-6i$, match the real and imaginary parts on both sides: $a-1=7$ and $b+3=-6$, giving $a=8$ and $b=-9$.

Pro Tip: When factoring completely, your final factors should be either linear or irreducible quadratics (those that can't be factored further with real numbers).

Working with Complex Numbers

Complex numbers open up a whole new world of solutions when real numbers aren't enough!

To add or subtract complex numbers, combine like terms separately. For real parts, add/subtract real parts; for imaginary parts, add/subtract imaginary parts. For example, $(3-3\sqrt{2}i) + (-4+6\sqrt{2}i) = -1+3\sqrt{2}i$. This works just like combining like terms in regular algebra.

Multiplying complex numbers uses the distributive property—multiply each term in the first expression by each term in the second. Remember that $i^2 = -1$, which simplifies your final result. For $(4+3i)(3-4i)$, you get $12-16i+9i-12i^2 = 12-7i+12 = 24-7i$.

Dividing complex numbers requires a special technique called rationalization. Multiply both numerator and denominator by the conjugate of the denominator (same expression but with the opposite sign for the imaginary part). This eliminates the imaginary part in the denominator. For $\frac{4+3i}{2-5i}$, multiply by $\frac{2+5i}{2+5i}$ to get $\frac{8+26i+15i^2}{4+25i^2} = \frac{-7+26i}{29} = \frac{-7+26i}{29}$.

Simplification Shortcut: When solving quadratics with complex solutions, the solutions always appear in complex conjugate pairs like $3+i\sqrt{215}$ and $3-i\sqrt{215}$.

Finding Rational Zeros

The Rational Zero Theorem gives you a list of possible rational zeros without having to test every number.

To find all possible rational zeros of a polynomial, list all factors of the constant term (p) and all factors of the leading coefficient (q). The possible rational zeros are $\pm\frac{p}{q}$. For example, in $4x^5 + 5x^4 - 3x^3 + 5x^2 - 7x - 10$, the constant term is -10 and the leading coefficient is 4, giving possible rational zeros like $\pm1, \pm2, \pm5, \pm10, \pm\frac{5}{2}, \pm\frac{5}{4}$.

After finding possible zeros, use your calculator to graph the function or use synthetic division to check each one. When you find an actual zero like 7 for $f(x) = 3x^3 - 22x^2 + 5x + 14$, use synthetic division to verify it and factor the polynomial as $(x-7)(3x^2-x-2)$.

You can factor further to get $(x-7)(3x+2)(x-1)$, giving all zeros: 7, $-\frac{2}{3}$, and 1. These zeros tell you exactly where the graph crosses the x-axis.

Test Strategy: On tests, try integers first when checking for rational zeros—they're usually easier to work with and are common solutions in classroom problems.

Complete Polynomial Factorization

Breaking down polynomials completely helps you understand their behavior and find all solutions.

When factoring polynomials, start by finding one zero using the Rational Zero Theorem and your calculator. For $f(x) = x^3 - 2x^2 - 5x + 6$, trying x = 1 and using synthetic division confirms it's a zero, giving $f(x) = (x-1)(x^2-x-6)$.

After finding the first factor, continue factoring the remaining polynomial. The quadratic factor $x^2-x-6$ can be factored as $(x-3)(x+2)$, giving the complete factorization $f(x) = (x-1)(x-3)(x+2)$. This tells you the function has zeros at x = 1, 3, and -2.

For higher-degree polynomials like $f(x) = x^4 - 4x^3 - 15x^2 + 18x$, you may need to repeat this process multiple times. Start by factoring out any common factors (like x) first. Then find one zero and continue factoring the resulting polynomial until you can't factor further.

Visual Connection: Each factor of the form $x-a$ corresponds to where the graph crosses the x-axis at the point a. The graph touches but doesn't cross at x=a when $x-a$ appears multiple times in the factorization.

Advanced Factoring Techniques

Some polynomials require special techniques to factor completely, especially when they have irrational or complex zeros.

When facing a difficult polynomial like $f(x) = 12x^3 - 183x^2 - 192$, try specific values that might simplify your work. After finding x = 4 is a zero and using synthetic division, you get $f(x) = (x-4)(12x^2+48x+36)$. The quadratic factor can be written as $12(x^2+4x+3)$, which factors further to $12(x+3)(x+1)$.

For polynomials with even powers and only even exponents, like $f(x) = x^4 - 21x^2 + 80$, try substituting $u = x^2$ to make factoring easier. This gives $u^2 - 21u + 80 = (u-16)(u-5) = (x^2-16)(x^2-5) = (x-4)(x+4)(x-\sqrt{5})(x+\sqrt{5})$.

Real-world problems often involve factoring polynomials to find dimensions or key values. For a box with volume $V(x) = x^3 - 14x^2 + 59x - 70$ where the length is $(x-7)$, use synthetic division to find the other factors: $(x-5)(x-2)$, giving width = 5 and height = 2.

Application Alert: In real-world problems, always check if your mathematical solution makes physical sense—negative dimensions aren't possible for physical objects!

Real-World Applications

Polynomial functions appear in many real-world situations, from designing boxes to modeling roller coaster tracks.

When creating an open box from a rectangular piece of cardboard, you cut equal squares from each corner and fold up the sides. If you start with a 9-by-11 inch piece and cut squares with side length x, the volume function is $V(x) = x(9-2x)(11-2x)$. This polynomial helps you find the maximum possible volume.

Polynomials also model physical situations like roller coaster designs. If $f(x) = x^4 - 8x^3 - 4x^2 + 128x - 192$ represents a roller coaster section, and you need to place a loading zone at one of the zeros, you must find all zeros. Starting with the known zero x = 6 and using synthetic division, you get $f(x) = (x-6)(x^3-2x^2-16x+32)$.

Further factoring gives $f(x) = (x-6)(x-2)(x-4)(x+4)$, showing that possible loading zone locations are at x = 2, 4, or 6 $not -4, as that wouldn't make physical sense$.

Real-World Context: When solving application problems, always consider the context to eliminate impossible solutions—like negative lengths or locations in physical space.