查看“︁初中數學（香港課程）/恆等式與公式/3”︁的源代码

平方差和完全平方可作因式分解
# <math>a^2-b^2\equiv (a+b)(a-b)</math>
# <math>a^2+2ab+b^2\equiv (a+b)^2</math>
# <math>a^2-2ab+b^2\equiv (a-b)^2</math>

{{MathHKQA|1=1
|2=因式分解
# <math>144-x^2</math>
# <math>25u^2-9v^2</math>
# <math>72p^2-98</math>
|sol=
# <math>
144-x^2=12^2-x^2=(12+x)(12-x)</math>
# <math>25u^2-9v^2=(5u+3v)(5u-3v)</math>
# <math>72p^2-98=2(36p^2-49)=2(6p+7)(6p-7)</math>
}}

{{MathHKQA|1=2
|2=因式分解
# <math>z^2-16z+64</math>
# <math>100a^2+180ab+81b^2</math>
# <math>n^3-2n^2+n</math>
|sol=
# <math>z^2-16z+64=z^2-2\cdot 8z+8^2=(z-8)^2</math>
# <math>100a^2+180ab+81b^2=(10a+9b)^2</math>
# <math>n^3-2n^2+n=n(n^2-2n+1)=n(n-1)^2</math>
}}

{{MathHKQA|1=3
|2=
# <math>(3y-5)^2-(y+4)^2</math>
# <math>49(z-6)^2+70(z-6)+25</math>
|sol=
# <math>\begin{align}
(3y-5)^2-(y+6)^2&=[(3y-5)+(y+4)][(3y-5)-(y+6)]\\
&=(3y-5+y+6)(3y-5-y-6)\\
&=(4y-1)(2y-11)
\end{align}</math>
# \begin{align}
49(z-6)^2+70(z-6)+25&=[7(z-6)-5]^2\\
&=(7z-42-5)^2\\
&=(7z-47)^2
\end{align}
}}

{{MathHKQA|1=4（綜合題）
|2=
# 因式分解<math>64l^2-176l+121</math>
# 從上題，因式分解<math>72l^4-99l^3-64l^2+176l-121</math>
# 從上題，化簡<math>\frac{3l}{72l^4-99l^3-64l^2+176l-121}\cdot \frac{56l-77}{6l^5}</math>
|sol=
# <math>64l^2-176l+121=(8l-11)^2</math>
# <math>\begin{align}
72l^4-99l^3-64l^2+176l-121&=9l^3(8l-11)-(64l^2-176l+121)\\
&=9l^3(8l-11)-(8l-11)^2\\
&=[9l^3-(8l-11)](8l-11)=(9l^3-8l+11)(8l-11)
\end{align}</math>
}}