Respuesta :
To factor both numerator and denominator in this rational expression we are going to substitute [tex]n^{2} [/tex] with [tex] x [/tex]; so [tex]n^{2} =x[/tex] and [tex]n ^{4} = x^{2} [/tex]. This way we can rewrite the expression as follows:
[tex] \frac{n^{4}-11n^{2} +30 }{n^{2}-7n^{2} +10 } = \frac{ x^{2} -11x+30}{ x^{2} -7x+10} [/tex]
Now we have two much easier to factor expressions of the form [tex]a x^{2} +bx+c[/tex]. For the numerator we need to find two numbers whose product is [tex]c[/tex] (30) and its sum [tex]b[/tex] (-11); those numbers are -5 and -6. [tex](-5)(-6)=30[/tex] and [tex]-5-6=-11[/tex].
Similarly, for the denominator those numbers are -2 and -5. [tex](-2)(-5)=10[/tex] and [tex]-2-5=-7[/tex]. Now we can factor both numerator and denominator:
[tex] \frac{ x^{2} -11x+30}{ x^{2} -7x+10} = \frac{(x-6)(x-5)}{(x-2)(x-5)} [/tex]
Notice that we have [tex](x-5)[/tex] in both numerator and denominator, so we can cancel those out:
[tex] \frac{x-6}{x-2} [/tex]
But remember than [tex]x= n^{2} [/tex], so lets replace that to get back to our original variable:
[tex] \frac{n^{2}-6 }{n^{2}-2 } [/tex]
Last but not least, the denominator of rational expression can't be zero, so the only restriction in the variable is [tex]n^{2} -2 \neq 0[/tex]
[tex] n^{2} \neq 2[/tex]
[tex] n \neq +or- \sqrt{2} [/tex]
[tex] \frac{n^{4}-11n^{2} +30 }{n^{2}-7n^{2} +10 } = \frac{ x^{2} -11x+30}{ x^{2} -7x+10} [/tex]
Now we have two much easier to factor expressions of the form [tex]a x^{2} +bx+c[/tex]. For the numerator we need to find two numbers whose product is [tex]c[/tex] (30) and its sum [tex]b[/tex] (-11); those numbers are -5 and -6. [tex](-5)(-6)=30[/tex] and [tex]-5-6=-11[/tex].
Similarly, for the denominator those numbers are -2 and -5. [tex](-2)(-5)=10[/tex] and [tex]-2-5=-7[/tex]. Now we can factor both numerator and denominator:
[tex] \frac{ x^{2} -11x+30}{ x^{2} -7x+10} = \frac{(x-6)(x-5)}{(x-2)(x-5)} [/tex]
Notice that we have [tex](x-5)[/tex] in both numerator and denominator, so we can cancel those out:
[tex] \frac{x-6}{x-2} [/tex]
But remember than [tex]x= n^{2} [/tex], so lets replace that to get back to our original variable:
[tex] \frac{n^{2}-6 }{n^{2}-2 } [/tex]
Last but not least, the denominator of rational expression can't be zero, so the only restriction in the variable is [tex]n^{2} -2 \neq 0[/tex]
[tex] n^{2} \neq 2[/tex]
[tex] n \neq +or- \sqrt{2} [/tex]
The simplified expression of the given rational function is [tex]\dfrac{n^2-6}{n^2-2}[/tex] and this can be determined by using the factorization method.
Given :
Rational Expression -- [tex]\dfrac{n^4-11n^2+30}{n^4-7n^2+10}[/tex]
The following calculation can be used to evaluate the given expression.
First, factorize the numerator separately.
[tex]=n^4-11n^2+30[/tex]
[tex]=n^4 -5n^2-6n^2+30[/tex]
[tex]=n^2(n^2-5)-6(n^2-5)[/tex]
[tex]=(n^2-5)(n^2-6)[/tex]
Now, factorize the denominator separately.
[tex]=n^4-7n^2+10[/tex]
[tex]=n^4-5n^2-2n^2+10[/tex]
[tex]=n^2(n^2-5)-2(n^2-5)[/tex]
[tex]=(n^2-5)(n^2-2)[/tex]
So, the rational expression becomes:
[tex]\dfrac{n^4-11n^2+30}{n^4-7n^2+10} = \dfrac{(n^2-5)(n^2-6)}{(n^2-5)(n^2-2)}[/tex]
[tex]=\dfrac{n^2-6}{n^2-2}[/tex]
In the rational function, the denominator never be zero that means:
[tex]n^2-2\neq 0[/tex]
[tex]n\neq -\sqrt{2} \; or \; \sqrt{2}[/tex]
For more information, refer to the link given below:
https://brainly.com/question/13101306
