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Methods of factorisation
In certain cases, for example when solving equations it can be useful to express an integral expression in product form.
Let us consider expression \latex{x^{3}+3x^{2}+2x}. As \latex{x} appears in all terms we can create a product one factor of which is \latex{x}, and the other factor will be formed so that the product of the factors is equal to the original expression:
\latex{x^{3}+3x^{2}+2x=\colorbox{#f3ded7}{$x$} \times x^{2}+\colorbox{#f3ded7}{$x$}\times 3x+\colorbox{#f3ded7}{$x$}\times 2=x\left(x^{2}+3x+2 \right)}.
Factoring out
If we create a product by dividing the terms of a sum by the same factor, it is called factoring out. For example:
\latex{12a^{4}-20a^{3}+8a^{2}=} \latex{\colorbox{#f3ded7}{ $4a^{2}$ } \times 3a^{2}-}\latex{\colorbox{#f3ded7}{ $4a^{2}$ }\times 5a+}\latex{\colorbox{#f3ded7}{ $4a^{2}$ }\times2=} \latex{=4a^{2}\left(3a^{2}-5a+2\right)};
\latex{6a^{2}b+9ab^{2}-12a^{2}b^{2}=3ab\left(2a+3b-4ab\right)};
\latex{16x^{3}y^{2}-24x^{2}y^{2}+32x^{2}y^{3}=8x^{2}y^{2}\left(2x-3+4y\right)}.
We keep applying the method till there are no more identical factors in the terms of the expression.
Factoring out by grouping
It can happen that we cannot factor out from each term, but by grouping cleverly we can still reach our goal. For example:
\latex{ax+bx+ay+by=\left(ax+bx\right)+\left(ay+by\right)=x\left(a+b\right)+y\left(a+b\right)=}
\latex{=\left(a+b\right)\left(x+y\right)};
\latex{=\left(a+b\right)\left(x+y\right)};
\latex{2bx+3a+2ax+3b=\left(2bx+2ax\right)+\left(3a+3b\right)=}
\latex{=2x\left(a+b\right)+3\left(a+b\right)=\left(a+b\right)\left(2x+3\right)};
\latex{=2x\left(a+b\right)+3\left(a+b\right)=\left(a+b\right)\left(2x+3\right)};
\latex{6ax-12xb+2ay-4by=\left(6ax-12xb\right)+\left(2ay-4by\right)=}
\latex{=6x\left(a-2b\right)+2y\left(a-2b\right)=\left(a-2b\right)\left(6x+2y\right)};
\latex{=6x\left(a-2b\right)+2y\left(a-2b\right)=\left(a-2b\right)\left(6x+2y\right)};
\latex{25x^{2}+10x+1=25x^{2}+5x+5x+1=5x\left(5x+1\right)+\left(5x+1\right)=}
\latex{=\left(5x+1\right)\left(5x+1\right)=\left(5x+1\right)^{2}}.
\latex{=\left(5x+1\right)\left(5x+1\right)=\left(5x+1\right)^{2}}.
Applying special identities
If we have a closer look at the expressions, we can often observe special identities in them. For example:
\latex{25x^{2}+10x+1=\left(5x+1\right)^2};
\latex{x^{3}+3x^{2}+3x+1=\left(x+1\right)^3};
\latex{4x^{2}-9=\left(2x-3\right)\left(2x+3\right)};
\latex{81x^{8}-16y^{4}=\left(9x^{4}+4y^{2}\right)\left(9x^{4}-4y^{2}\right)=\left(9x^{4}+4y^{2}\right)\left(3x^{2}+2y\right)\left(3x^{2}-2y\right)}.
As the last example shows it is worth checking the resulting factors when factorising, because it is possible that they can be transformed further.
Finally let us have some compound exercises where the special products cannot be seen immediately but can be formed. For example:
\latex{3x^{3}-3xy^{2}=3x\left(x^{2}-y^{2}\right)=3x\left(x+y\right)\left(x-y\right)};
\latex{4ax^{2}-12ax+9a=a\left(4x^{2}-12x+9\right)=a\left(2x-3\right)^2};
\latex{48b^{3}+120b^{2}+75b=3b\left(16b^{2}+40b+25\right)=3b\left(4b+5\right)^2};
\latex{x^{2}+6x+5=\left(x^{2}+6x+9 \right)-9+5=\left(x+3\right)^2-4=\left(x+3\right)^2-2^{2}=}
\latex{=\left(x+3+2\right)\left(x+3-2\right)=\left(x+5\right)\left(x+1\right)}.
\latex{=\left(x+3+2\right)\left(x+3-2\right)=\left(x+5\right)\left(x+1\right)}.
Exercises
{{exercise_number}}. Factorise the expressions below.
- \latex{12x^{3}-8x^{2}+4x}
- \latex{6a^{3}b-8a^{2}b^{2}}
- \latex{20x^{2}y-30xy^{2}}
- \latex{7x^{2}y^{3}-14x^{3}y^{3}+21x^{2}y^{4}}
- \latex{18a^{7}b^{4}+6a^{5}b^{7}+30a^{10}b^{3}}
- \latex{24a^{2}b^{3}-16a^{3}b^{2}}
{{exercise_number}}. Factorise the expressions below.
- \latex{ax-bx-ay+by}
- \latex{6ax+12x+2ay+4y}
- \latex{4ax-28a-bx+7b}
- \latex{15ax-10ay+6bx-4by}
- \latex{12ax-2bx+18ay-3by}
- \latex{6xy+4x-3y-2}
- \latex{6ax^{2}-9x^{2}+2a-3}
- \latex{8a^{2}x-b^{2}y+4a^{2}y-2b^{2}x}
{{exercise_number}}. Factorise the expressions below.
- \latex{64x^{2}+9-48x}
- \latex{121+88x+16x^{2}}
- \latex{9a^{2}-49b^{2}}
- \latex{\frac{4}{9}x^{2}-y^{2}}
- \latex{49a^{4}+28a^{2}b+4b^{2}}
- \latex{\frac{1}{2}mv_{2}^2 - \frac{1}{2}mv_{1}^2}
- \latex{36a^{6}+25b^{4}-60a^{3}b^{2}}
- \latex{\frac{9}{49}x^{2}-\frac{4}{7}xy+\frac{4}{9}y^{2}}
- \latex{a^{16}-1}
{{exercise_number}}. Factorise the expressions below.
- \latex{45x^{2}-120x+80}
- \latex{3a^{6}+18a^{4}b+27a^{2}b^{2}}
- \latex{8a^{2}b^{3}-8a^{4}b^{2}+2a^{6}b}
- \latex{x^{2}-4x-21}
- \latex{2x^{2}+8x-10}
- \latex{3x^{4}+13x^{2}+12}
{{exercise_number}}. Factorise the expressions below.
- \latex{x^{4}+x^{2}+1}
- \latex{x^{4}+4}
- \latex{x^{8}+4}
Puzzle
The difference of the squares of two natural numbers is \latex{225}. Give all such natural numbers.