How to Solve Higher Degree Polynomials

Co-authored by Jake Adams

Last Updated: July 22, 2024 Fact Checked

Solving a higher degree polynomial has the same goal as a quadratic or a simple algebra expression: factor it as much as possible, then use the factors to find solutions to the polynomial at y = 0. There are many approaches to solving polynomials with an $x^{3}$ term or higher. You may need to use several before you find one that works for your problem.

Method 1

Method 1 of 2:

Recognizing Factors

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1
Factor out common factors from all terms. If every term in the polynomial has a common factor, factor it out to simplify the problem. This is not possible with all polynomials, but it's a good approach to check first. ^{[1]
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- Example 1: Solve for x in the polynomial $2x^{{3}}+12x^{{2}}+16x=0$ .
  Each term is divisible by 2x, so factor it out:
  $(2x)(x^{{2}})+(2x)(6x)+(2x)(8)=0$
  $=(2x)(x^{{2}}+6x+8)$
  Now solve the quadratic equation using the quadratic formula or factoring:
  $(2x)(x+4)(x+2)=0$
  The solutions are at 2x = 0, x+4=0, and x+2=0.
  The solutions are x=0, x=-4, and x=-2 .
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2
Identify polynomials that act like a quadratic. You likely already know how to solve second degree polynomials, in the form $ax^{{2}}+bx+c$ . You can solve some higher-degree polynomials the same way, if they're in the form $ax^{{2n}}+bx^{{n}}+c$ . Here are a couple examples:
- Example 2: $3x^{{4}}+4x^{{2}}-4=0$
  Let $a=x^{{2}}$ :
  $3a^{{2}}+4a-4=0$
  Solve the quadratic using any method:
  $(3a-2)(a+2)=0$ so a = -2 or a = 2/3
  Substitute $x^{2}$ for a: $x^{{2}}=-2$ or $x^{{2}}=2/3$
  x = ±√(2/3) . The other equation, $x^{{2}}=-2$ , has no real solution. (If using complex numbers, solve as x = ±i√2 ).
- Example 3: $x^{{5}}+7x^{{3}}-9x=0$ does not follow this pattern, but notice you can factor out an x:
  $(x)(x^{{4}}+7x^{{2}}-9)=0$
  You can now treat $x^{{4}}+7x^{{2}}-9$ as a quadratic, as shown in example 2.
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3
Factor sums or differences of cubes. These special cases look difficult to factor, but have properties which make the problem much easier:
- Sum of cubes: A polynomial in the form $a^{{3}}+b^{{3}}$ factors to $(a+b)(a^{{2}}-ab+b^{{2}})$ . ^{[2]
  X
  Research source}
- Difference of cubes: A polynomial in the form $a^{{3}}-b^{{3}}$ factors to $(a-b)(a^{{2}}+ab+b^{{2}})$ . ^{[3]
  X
  Research source}
- Note that the quadratic portion of the result is not factorable. ^{[4]
  X
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- Note that $x^{6}$ , $x^{9}$ , and x to any power divisible by 3 all fit these patterns.
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4
Look for patterns to find other factors. Polynomials that do not look like the examples above may not have any obvious factors. But before you try the methods below, try looking for a two-term factor (such as "x+3"). Grouping terms in different orders and factoring out part of the polynomial may help you find one. ^{[5]
X
Research source} This is not always a feasible approach, so don't spend too much time trying if no common factor seems likely.
- Example 4: $-3x^{{3}}-x^{{2}}+6x+2=0$
  This has no obvious factor, but you can factor the first two terms and see what happens:
  $(-x^{{2}})(3x+1)+6x+2=0$
  Now factor the last two terms (6x+2), aiming for a common factor:
  $(-x^{{2}})(3x+1)+(2)(3x+1)=0$
  Now rewrite this using the common factor, 3x+1:
  $(3x+1)(-x^{{2}}+2)=0$

Method 2

Method 2 of 2:

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Question

How can I tell if a function is even or odd?

Jake Adams
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Expert Answer

Functions are typically categorized as either even or odd based on their behavior concerning the X-axis and the parity of their polynomial values. However, a function may be considered neither even nor odd if it represents a constant. For example, in the case of Y equals 4, the graph forms a horizontal line at Y equals 4, creating a unique Y value for each X input and satisfying the vertical line test. Despite having no repeated Y values for any X input, a constant function like Y equals 4 is neither even nor odd because it doesn't intersect the X-axis. This distinction arises from the fact that the function neither exhibits symmetry about the Y-axis (even) nor about the origin (odd).

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In JEE (adv) 2011, a question was to find the number of roots of x^4 - 4^3 + 12x^2 +1 = 0; although I have the solution, how would I solve an equation like this?

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If you try the rational root test, you find that the equation has no rational roots. Therefore, you will have to use numerical methods to find the other roots. You could try the Newton-Raphson Method for the irrational roots.

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How do I solve x^5 - 4x^4 + x³ + 10x² - 4x - 8?

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First of all, there is no meaningful simplification of that expression. Secondly, remember that in order to "solve" such an expression, it must be presented in the form of an equation.

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Tips

Descartes' Rule of Signs won't tell you the solution, but it can predict how many unique, real solutions there are. Follow these steps to find out whether you've found all possible solutions: ^{[12]
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Research source}
- Arrange the polynomial from highest degree term to lowest:
  $x^{5}-x^{4}-2x^{2}+x+1$
- Ignore the terms and write just their signs (positive or negative)
  +--++
- Count the number of times the signs changed from + to - or vice versa, moving left to right:
  The sequence +--++ shifts signs 2 times.
- The number of real solutions is either equal to that number, or equal to that number minus 2 n , where n is an integer.
  In this example, there may be 2 solutions, or there may be 0.
  In another hypothetical problem where the terms change signs seven times, the number of solutions could be 7, 5, 3, or 1.
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Cubic and quartic formulas exist similar to the quadratic formula, but are much more complicated and are not often used except by computer. Polynomials of degree 5 and higher have no general solution using simple algebraic techniques, but some examples can be factored using the approaches above.

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The terms roots , zeros , and solutions all refer to the values of x that make f(x) = 0. They may be used interchangeably.

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Warnings

If you get an imaginary root (and you are working with a problem where imaginary roots matter), don't forget that there will be a zero at that number and its complex conjugate. If (x-3i) is a root, then so is (x+3i).

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How to Solve Higher Degree Polynomials

Steps

Recognizing Factors

Rational Roots and Synthetic Division

Testing for Rational Roots

Synthetic Division

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