Question

Simplifying rational expressions x^2-16/x^2+6x+8 = x-4/x+2 x^2-x-6/x^2-3x-10 = x-3/x-5

Answer 1

1)

$\dfrac{x^2-16}{x^2+6x+8}$

Decompose the numerator and denominator into multipliers

To simplify the numerator we use the formula of difference of squares

$x^2-y^2=(x-y)(x+y)$

$x^2-16=(x-4)(x+4)$

To decompose the denominator into multipliers solve the square equation

$x^2+6x+8=0\\D=6^2-4*8=4=2^2\\x_1=\dfrac{-6+2}{2} =-2\\x_2=\dfrac{-6-2}{2} =-4$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

$(x-(-2))(x-(-4))=(x+2)(x+4)$

After simplifying the numerator and denominator we get a fraction

$\dfrac{(x-4)(x+4)}{(x+2)(x+4)}=\dfrac{x-4}{x+2}$, so

$\dfrac{x^2-16}{x^2+6x+8}=\dfrac{(x-4)(x+4)}{(x+2)(x+4)}=\dfrac{x-4}{x+2}$

2)

$\dfrac{x^2-x-6}{x^2-3x-10}$

Decompose the numerator and denominator into multipliers

To decompose the numerator into multipliers solve the square equation

$x^2-x-6=0\\D=(-1)^2-4*(-6)=25=5^2\\x_1=\dfrac{1+5}{2} =3\\x_2=\dfrac{1-5}{2} =-2$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

$(x-3)(x-(-2))=(x-3)(x+2)$

To decompose the denominator into multipliers solve the square equation

$x^2-3x-10=0\\D=(-3)^2-4*(-10)=49=7^2\\x_1=\dfrac{3+7}{2} =5\\x_2=\dfrac{3-7}{2} =-2$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

$(x-5)(x-(-(-2))=(x-5)(x+2)$

After simplifying the numerator and denominator we get a fraction

$\dfrac{(x-3)(x+2)}{(x-5)(x+2)}=\dfrac{x-3}{x-5}$, so

$\dfrac{x^2-x-6}{x^2-3x-10}=\dfrac{(x-3)(x+2)}{(x-5)(x+2)}=\dfrac{x-3}{x-5}$

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