Question

application: 1. simplify the following expression. state the restrictions. $\frac{8x}{2x - 4}cdot\frac{x^{2}+5x + 6}{x^{2}-3x + 2}cdot\frac{x + 3}{2x - 2}$ 7 thinking: 1. determine the area of the triangle in simplified form. state the restrictions. hint: $a=\frac{bh}{2}$ 6 $\frac{5x - 35}{x + 3}$ $\frac{4x^{2}}{x^{2}-16x + 63}$

Explanation:

Step1: Factor the expressions

• Factor the numerators and denominators:
• For $\frac{8x}{2x - 4}=\frac{8x}{2(x - 2)}=\frac{4x}{x - 2}$;
• $x^{2}+5x + 6=(x + 2)(x+3)$;
• $x^{2}-3x + 2=(x - 1)(x - 2)$;
• $\frac{x + 3}{2x-2}=\frac{x + 3}{2(x - 1)}$.
• The original expression $\frac{8x}{2x - 4}\cdot\frac{x^{2}+5x + 6}{x^{2}-3x + 2}\cdot\frac{x + 3}{2x-2}$ becomes $\frac{4x}{x - 2}\cdot\frac{(x + 2)(x + 3)}{(x - 1)(x - 2)}\cdot\frac{x + 3}{2(x - 1)}$.

Step2: Multiply the fractions

• Multiply the numerators together and denominators together: $\frac{4x(x + 2)(x + 3)(x + 3)}{2(x - 1)(x - 1)(x - 2)(x - 2)}=\frac{2x(x + 2)(x + 3)^{2}}{(x - 1)^{2}(x - 2)^{2}}$.

Step3: Find the restrictions

• Set the denominators of the original - non - simplified fractions equal to zero.
• For $2x-4 = 0$, $x = 2$; for $x^{2}-3x + 2=0$, $(x - 1)(x - 2)=0$, $x = 1$ or $x = 2$; for $2x-2 = 0$, $x = 1$.
• The restrictions are $x

eq1,x
eq2$.

Step4: For the triangle area problem

• First, factor the expressions in the base and height of the triangle.
• The height $h=\frac{5x-35}{x + 3}=\frac{5(x - 7)}{x + 3}$, and the base $b=\frac{4x^{2}}{x^{2}-16x + 63}=\frac{4x^{2}}{(x - 7)(x - 9)}$.
• Using the area formula $A=\frac{1}{2}bh$, we have $A=\frac{1}{2}\cdot\frac{5(x - 7)}{x + 3}\cdot\frac{4x^{2}}{(x - 7)(x - 9)}$.
• Cancel out the common factor $(x - 7)$: $A=\frac{1}{2}\cdot\frac{5}{x + 3}\cdot\frac{4x^{2}}{x - 9}=\frac{10x^{2}}{(x + 3)(x - 9)}$.
• The restrictions are found by setting the denominators of the original non - simplified height and base expressions equal to zero. For $x+3 = 0$, $x=-3$; for $(x - 7)(x - 9)=0$, $x = 7$ or $x = 9$. So the restrictions are $x

eq - 3,x
eq7,x
eq9$.

Answer:

• For the first part: Simplified form is $\frac{2x(x + 2)(x + 3)^{2}}{(x - 1)^{2}(x - 2)^{2}}$, restrictions are $x

eq1,x
eq2$.

• For the second part: Simplified form of the triangle area is $\frac{10x^{2}}{(x + 3)(x - 9)}$, restrictions are $x

eq - 3,x
eq7,x
eq9$.