The factors of 32 are 1, 2, 4, 8, 16, and 32 Both “1” and the number you’re factoring are always factors. So, the factors of a small number, like 3, would simply be 1 and 3. Factors are only the perfectly divisible numbers, or “whole” numbers. You could divide 32 by 3. 564, or 21. 4952, but this won’t lead to a factor, just another decimal.
3t+6{\displaystyle 3t+6} → 6+3t{\displaystyle 6+3t} 3x4+9x2{\displaystyle 3x^{4}+9x^{2}} → 9x2+3x4{\displaystyle 9x^{2}+3x^{4}} x2−2{\displaystyle x^{2}-2} → −2+x2{\displaystyle -2+x^{2}} Note how the negative sign stays in front of the 2. If a term is subtracted, just keep the negative in front of it.
Practice Problem:3t+6{\displaystyle 3t+6}. Factors of 3: 1, 3 Factors of 6: 1, 2, 3, 6. The greatest common factor is 3.
Factors of 3: 1, 3 Factors of 6: 1, 2, 3, 6. The greatest common factor is 3.
Practice Problem:3t+6{\displaystyle 3t+6}. Find greatest common factor: 3 Remove factor from both terms:3t3+63=t+2{\displaystyle {\frac {3t}{3}}+{\frac {6}{3}}=t+2}
Practice Problem:3t+6{\displaystyle 3t+6} Find greatest common factor: 3 Remove factor from both terms:3t3+63=t+2{\displaystyle {\frac {3t}{3}}+{\frac {6}{3}}=t+2} Multiple factor by new expression: 3(t+2){\displaystyle 3(t+2)} Final Factored Answer: 3(t+2){\displaystyle 3(t+2)}
Reorganize terms:18+12t{\displaystyle 18+12t} Find greatest common denominator: 6{\displaystyle 6} Remove factor from both terms:18t6+12t6=3+2t{\displaystyle {\frac {18t}{6}}+{\frac {12t}{6}}=3+2t} Multiple factor by new expression: 6(3+2t){\displaystyle 6(3+2t)} Check Answer: (6∗3)+(6∗2t)=18+12t{\displaystyle (63)+(62t)=18+12t}
Practice Problem: 5y−2y2=−3y{\displaystyle 5y-2y^{2}=-3y} Remember that binomials must only have two terms. If there are more than two terms you can learn to solve polynomials instead.
Practice Problem: 5y−2y2=−3y{\displaystyle 5y-2y^{2}=-3y} Set to Zero: 5y−2y2+3y=−3y+3y{\displaystyle 5y-2y^{2}+3y=-3y+3y} 8y−2y2=0{\displaystyle 8y-2y^{2}=0}
8y−2y2=0{\displaystyle 8y-2y^{2}=0}
Practice Problem: 5y−2y2=−3y{\displaystyle 5y-2y^{2}=-3y} Set to Zero: 8y−2y2=0{\displaystyle 8y-2y^{2}=0} Factor: 2y(4−y)=0{\displaystyle 2y(4-y)=0}
Practice Problem: 5y−2y2=−3y{\displaystyle 5y-2y^{2}=-3y} Set to Zero: 8y−2y2+3y=0{\displaystyle 8y-2y^{2}+3y=0} Factor: 2y(4−y)=0{\displaystyle 2y(4-y)=0} Set both parts to 0: 2y=0{\displaystyle 2y=0} 4−y=0{\displaystyle 4-y=0}
2y=0{\displaystyle 2y=0} 4−y=0{\displaystyle 4-y=0}
2y=0{\displaystyle 2y=0} 2y2=02{\displaystyle {\frac {2y}{2}}={\frac {0}{2}}} y = 0 4−y=0{\displaystyle 4-y=0} 4−y+y=0+y{\displaystyle 4-y+y=0+y} y = 4
5(0)−2(0)2=−3(0){\displaystyle 5(0)-2(0)^{2}=-3(0)} 0+0=0{\displaystyle 0+0=0} 0=0{\displaystyle 0=0} This answer is correct 5(4)−2(4)2=−3(4){\displaystyle 5(4)-2(4)^{2}=-3(4)} 20−32=−12{\displaystyle 20-32=-12} −12=−12{\displaystyle -12=-12} This answer is also correct.
2t+t2{\displaystyle 2t+t^{2}} can be factored, because both terms contain a t. Your final answer would be t(2+t){\displaystyle t(2+t)} You can even pull out multiple variables at once. For example, in x2+x4{\displaystyle x^{2}+x^{4}} both terms contain the same x2{\displaystyle x^{2}}. You can factor to x2(1+x2){\displaystyle x^{2}(1+x^{2})}
Original Problem: 6+2x+14+3x{\displaystyle 6+2x+14+3x} Reorganize terms: 2x+3x+14+6{\displaystyle 2x+3x+14+6} Combine like terms: 5x+20{\displaystyle 5x+20} Find greatest common factor: 5(x)+5(4){\displaystyle 5(x)+5(4)} Factor: 5(x+4){\displaystyle 5(x+4)}
Difference of perfect squares formula: a2−b2=(a+b)(a−b){\displaystyle a^{2}-b^{2}=(a+b)(a-b)} Practice Problem: 4x2−9{\displaystyle 4x^{2}-9} Find square roots: 4x2=2x{\displaystyle {\sqrt {4x^{2}}}=2x} 9=3{\displaystyle {\sqrt {9}}=3} Plug squares into formula: 4x2−9=(2x+3)(2x−3){\displaystyle 4x^{2}-9=(2x+3)(2x-3)}
4x2=2x{\displaystyle {\sqrt {4x^{2}}}=2x} 9=3{\displaystyle {\sqrt {9}}=3}
Difference of perfect cubes formula: a3−b3=(a−b)(a2+ab+b2){\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})} Practice Problem: 8x3−27{\displaystyle 8x^{3}-27} Find cubed roots: 8x33=2x{\displaystyle {\sqrt[{3}]{8x^{3}}}=2x} 273=3{\displaystyle {\sqrt[{3}]{27}}=3} Plug cubes into formula: 8x3−27=(2x−3)(4x2+6x+9){\displaystyle 8x^{3}-27=(2x-3)(4x^{2}+6x+9)}[16] X Research source
8x33=2x{\displaystyle {\sqrt[{3}]{8x^{3}}}=2x} 273=3{\displaystyle {\sqrt[{3}]{27}}=3}
Sum of perfect cubes formula: a3+b3=(a+b)(a2−ab+b2){\displaystyle a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})} Practice Problem: 8x3−27{\displaystyle 8x^{3}-27} Find cubed roots: 8x33=2x{\displaystyle {\sqrt[{3}]{8x^{3}}}=2x} 273=3{\displaystyle {\sqrt[{3}]{27}}=3} Plug cubes into formula: 8x3−27=(2x+3)(4x2−6x+9){\displaystyle 8x^{3}-27=(2x+3)(4x^{2}-6x+9)}[18] X Research source
8x33=2x{\displaystyle {\sqrt[{3}]{8x^{3}}}=2x} 273=3{\displaystyle {\sqrt[{3}]{27}}=3}
