Rekenen met variabelen *: Notable Products
The difference of two squares
The following formula is often used to factorize an expression in two terms from right to left
Notable product \[(a-b)(a+b)=a^2-b^2\]
The derivation of the sum formula is the following: \[ \begin{array}{rclcl}(a-b)(a+b)&=&a^2-ba+ab-b^2&\phantom{x}&\color{blue}{\text{expanding brackets}}\\ &=&a^2-ab+ab-b^2 &&\color{blue}{ba=ab}\\ &=&a^2-b^2&&\color{blue}{\text{adding up the terms}}\end {array}\]
The formula is used to make the denominator of a fraction free from roots like in the next case.
When #a#, #b#, #c#, #d#, and #e# are integers having #a\gt0# and #d^2\ne e^2a#, then
\[\frac{b+c\sqrt{a}}{d+e\sqrt{a}}=\frac{(b\cdot d-a\cdot c\cdot e) +(c\cdot d-b\cdot e)\sqrt{a}}{d^2-e^2a}\tiny.\]
After all, \[ \begin{array}{rcl}\frac{b+c\sqrt{a}}{d+e\sqrt{a}}&=&\frac{\left(b+c\sqrt{a}\right)\cdot\left(d-e\sqrt{a}\right)}{\left(d+e\sqrt{a}\right)\cdot\left(d-e\sqrt{a}\right)}\\&&\phantom{space}\color{blue}{\text{numerator and denominator are multiplied by }d-e\sqrt{a}}\\&=&\frac{(b\cdot d-a\cdot c\cdot e) +(c\cdot d-b\cdot e)\sqrt{a}}{d^2-e^2a}\\&&\phantom{space}\color{blue}{\text{brackets are removed with the notable product}}\end {array}\]
It is not necessary to memorize the formula. It is better to remember the method: if the denominator is #d+e\sqrt{a}#, then the roots can be eliminated by multiplying the notable product by #d-e\sqrt{a}#. In order to make the fraction correspond with the same number, the numerator should also be multiplied by #d-e\sqrt{a}#.
Solution: \[ \begin{array}{rclcl}{16}{y}^2-361&=&(4y)^2-19^2&&\color{blue}{\text{recognize the square}}\\&=&(4y+19)(4y-19)&&\color{blue}{\text{factorize}}\end {array}\]
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