Example 1: Solve for x in the polynomial 2x3+12x2+16x=0{\displaystyle 2x^{3}+12x^{2}+16x=0}. Each term is divisible by 2x, so factor it out:(2x)(x2)+(2x)(6x)+(2x)(8)=0{\displaystyle (2x)(x^{2})+(2x)(6x)+(2x)(8)=0}=(2x)(x2+6x+8){\displaystyle =(2x)(x^{2}+6x+8)}Now solve the quadratic equation using the quadratic formula or factoring:(2x)(x+4)(x+2)=0{\displaystyle (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.
Example 2: 3x4+4x2−4=0{\displaystyle 3x^{4}+4x^{2}-4=0}Let a=x2{\displaystyle a=x^{2}}:3a2+4a−4=0{\displaystyle 3a^{2}+4a-4=0}Solve the quadratic using any method:(3a−2)(a+2)=0{\displaystyle (3a-2)(a+2)=0} so a = -2 or a = 2/3Substitute x2{\displaystyle x^{2}} for a: x2=−2{\displaystyle x^{2}=-2} or x2=2/3{\displaystyle x^{2}=2/3}x = ±√(2/3). The other equation, x2=−2{\displaystyle x^{2}=-2}, has no real solution. (If using complex numbers, solve as x = ±i√2). Example 3: x5+7x3−9x=0{\displaystyle x^{5}+7x^{3}-9x=0} does not follow this pattern, but notice you can factor out an x:(x)(x4+7x2−9)=0{\displaystyle (x)(x^{4}+7x^{2}-9)=0}You can now treat x4+7x2−9{\displaystyle x^{4}+7x^{2}-9} as a quadratic, as shown in example 2.
Sum of cubes: A polynomial in the form a3+b3{\displaystyle a^{3}+b^{3}} factors to (a+b)(a2−ab+b2){\displaystyle (a+b)(a^{2}-ab+b^{2})}. [2] X Research source Difference of cubes: A polynomial in the form a3−b3{\displaystyle a^{3}-b^{3}} factors to (a−b)(a2+ab+b2){\displaystyle (a-b)(a^{2}+ab+b^{2})}. [3] X Research source Note that the quadratic portion of the result is not factorable. [4] X Research source Note that x6{\displaystyle x^{6}}, x9{\displaystyle x^{9}}, and x to any power divisible by 3 all fit these patterns.
Example 4: −3x3−x2+6x+2=0{\displaystyle -3x^{3}-x^{2}+6x+2=0}This has no obvious factor, but you can factor the first two terms and see what happens:(−x2)(3x+1)+6x+2=0{\displaystyle (-x^{2})(3x+1)+6x+2=0}Now factor the last two terms (6x+2), aiming for a common factor:(−x2)(3x+1)+(2)(3x+1)=0{\displaystyle (-x^{2})(3x+1)+(2)(3x+1)=0}Now rewrite this using the common factor, 3x+1:(3x+1)(−x2+2)=0{\displaystyle (3x+1)(-x^{2}+2)=0}
The root of a polynomial is the value of x for which y = 0. Knowing a root c also gives you a factor of the polynomial, (x - c).
Example: The polynomial 2x3+x2−12x+9{\displaystyle 2x^{3}+x^{2}-12x+9} has the constant term 9. Its factors are 1, 3, and 9.
Example (cont. ): 2x3+x2−12x+9{\displaystyle 2x^{3}+x^{2}-12x+9} has a leading coefficient of 2. Its factors are 1 and 2.
Example (cont. ): Any rational roots of this polynomial are in the form (1, 3, or 9) divided by (1 or 2). Possibilities include ±1/1, ±1/2, ±3/1, ±3/2, ±9/1, or ±9/2. Don’t forget the “±”: each of these possibilities could be positive or negative.
Example: (1/1=1) is a possible root. If it turns out to be an actual root, plugging it into the polynomial should result in zero. 2(1)3+(1)2−12(1)+9=2+1−12+9=0{\displaystyle 2(1)^{3}+(1)^{2}-12(1)+9=2+1-12+9=0}, so 1 is confirmed to be a root. This means the polynomial has the factor (x-1). If none of the possibilities work out, the polynomial has no rational roots and cannot be factored.
Note: You may need to insert terms with a coefficient of zero. For example, rewrite the polynomial x3+2x{\displaystyle x^{3}+2x} as x3+0x2+2x+0{\displaystyle x^{3}+0x^{2}+2x+0}. Example (cont. ): The rational roots test above told us that the polynomial 2x3+x2−12x+9{\displaystyle 2x^{3}+x^{2}-12x+9} has the root 1. Write the root 1, followed by a vertical line, followed by the coefficients of the polynomial:(1|21−129){\displaystyle {\begin{pmatrix}1|&2&1&-12&9\end{pmatrix}}}
Example (cont. ): Carry the 2 down to the answer line:(1|21−1292){\displaystyle {\begin{pmatrix}1|&2&1&-12&9\\&2\end{pmatrix}}}
Example (cont. ): Multiply the 2 by the root, 1, to get 2 again. Write this 2 in the following column, but on the second row instead of the answer line:(1|21−12922){\displaystyle {\begin{pmatrix}1|&2&1&-12&9\&&2\&2\end{pmatrix}}}
Example (cont. ): 1 + 2 = 3(1|21−129223){\displaystyle {\begin{pmatrix}1|&2&1&-12&9\&&2\&2&3\end{pmatrix}}}
Example (cont. ): 1 x 3 = 3:(1|21−1292323){\displaystyle {\begin{pmatrix}1|&2&1&-12&9\&&2&3\&2&3\end{pmatrix}}}
Example (cont. ): -12 + 3 = -9:(1|21−1292323−9){\displaystyle {\begin{pmatrix}1|&2&1&-12&9\&&2&3\&2&3&-9\end{pmatrix}}}
Example (cont. ): Multiply -9 by the root 1, write the answer under the final column, then confirm that the sum of the final column is zero:(1|21−12923−923−90){\displaystyle {\begin{pmatrix}1|&2&1&-12&9\&&2&3&-9\&2&3&-9&0\end{pmatrix}}}
Example (cont. ): The answer line is 2 3 -9 0, but you can ignore the final zero. Since the first term of the original polynomial included an x3{\displaystyle x^{3}}, the first term of your answer is one degree lower: x2{\displaystyle x^{2}}. Therefore, the first term is 2x2{\displaystyle 2x^{2}}Repeat this process to get the answer 2x2+3x−9{\displaystyle 2x^{2}+3x-9}. You have now factored 2x3+x2−12x+9{\displaystyle 2x^{3}+x^{2}-12x+9} into (x−1)(2x2+3x−9){\displaystyle (x-1)(2x^{2}+3x-9)}.
Remember, to start the synthetic division method, you’ll need to know one root already. Use the rational roots test again to get this. If none of the rational root possibilities check out, the expression cannot be factored. Example (cont. ) You’ve found the factors (x−1)(2x2+3x−9){\displaystyle (x-1)(2x^{2}+3x-9)}, but the second factor can be broken down further. Try the quadratic equation, traditional factoring, or synthetic division. The final answer is (x−1)(x+3)(2x−3){\displaystyle (x-1)(x+3)(2x-3)}, so the roots of the polynomial are x = 1, x = -3, and x = 3/2.
