Remember, factoring is finding out what numbers can divide into the whole. The expression x4{\displaystyle x^{4}} is another way of saying x∗x∗x∗x{\displaystyle x*x*x*x}. This means you can factor out each x if the other term has one as well. Treat variables no different from a normal number. For example:   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})} Take, for example, the expression 6+2x+14+3x{\displaystyle 6+2x+14+3x}. This may seem like it has four terms, but look closely and you'll realize there are really only two. You can add like terms, and since both the 6 and 14 have no variable, and the 2x and 3x share the same variable, these can both be combined. Factoring is then easy:   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)} " A perfect square is a number whose square root is a whole number, like 9{\displaystyle 9} (3∗3){\displaystyle (3*3)}, x2{\displaystyle x^{2}} (x∗x){\displaystyle (x*x)}, or even 144t2{\displaystyle 144t^{2}} (12t∗12t){\displaystyle (12t*12t)} If your binomial is a subtraction problem with two perfect squares, like a2−b2{\displaystyle a^{2}-b^{2}}, you can simply plug them into this formula:   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)} " Just like the perfect squares, this is a simple formula for when you have two cubed terms subtracted by each other. For example, a3−b3{\displaystyle a^{3}-b^{3}}. Just like before, you simply find the cubed root of each, plugging them into a formula:   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)} Unlike the difference of perfect squares, you can easily find added cubes too, like a3+b3{\displaystyle a^{3}+b^{3}}, with a simple formula. It's almost the exact same as above, just with some pluses and minuses flipped. The formula is just as easy as the other two, and all you have to do is recognize the two cubes in the problem to use it:   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)}
تذكر أن المتغيرات تعد عوامل كذلك، حتى مع الأسس. تعرف على ثنائيات الحدود غير المبسطة من خلال جمع الحدود المتشابهة. تعرف على القانون الخاص "بالفرق بين مربعين كاملين". تعلم قانون تحليل "الفرق بين مكعبين". اعرف أن لمجموع المكعبات الكاملة كذلك قانونًا خاصًا.