How do you simplify and find the excluded value of #(2x^2+6x+4)/(4x^2-12x-16)#?

Question

Abigail Allen · Accepted Answer

#1/2* (x+2)/(x-4)# #[2(x^2+3x+2)]/[4(x^2-3x-4)] = 1/2*[x^2+(x+2x)+2]/[x^2-(4x-x)-4]# #rArr 1/2*[x^2+x+2x+2]/[x^2-4x+x-4]# #rArr 1/2*[x(x+1)+2(x+1)]/[x(x-4)+1(x-4)]# #rArr 1/2* [(x+1)(x+2)]/[(x+1)(x-4)]# #rArr 1/2*(x+2)/(x-4)#

Kennedy Adams · Answer

#= (x+2)/(2(x-4))#

This means that #x!=4# The first step in algebraic fractions is to factor as far as possible: #(2x^2+6x+4)/(4x^2-12x-16)" "(larr"common factor")/(larr "common factor")# #= (2(x^2+3x+2))/(4(x^2-3x-4))(larr"quadratic trinomial")/(larr "quadratic trinomial")# #= (2(x+2)(x+1))/(4(x-4)(x+1))# #= (cancel2(x+2)cancel((x+1)))/(cancel4^2(x-4)cancel((x+1)))" "larr# cancel like factors #= (x+2)/(2(x-4))#

Jace Anderson · Answer

undefined To simplify the expression (2x^2+6x+4)/(4x^2-12x-16), we can first factor both the numerator and denominator. The numerator can be factored as 2(x+2)(x+1), and the denominator can be factored as 4(x-2)(x+2).

Next, we can cancel out any common factors between the numerator and denominator. In this case, we can cancel out the (x+2) term.

After canceling out the common factor, the simplified expression becomes 2(x+1)/(4(x-2)).

To find the excluded value, we need to determine the value(s) of x that would make the denominator equal to zero. In this case, the excluded value is x = 2, as it would make the denominator 4(2-2) = 0.