Asked by Sandra
4 This is hard for me to understand how to work them...
What is the simplified form of the product
[(x^2 + 8x + 15)/(x-4)] * [(x^2 – 16) / 2x + 6)]
What is the simplified form of the product
[(x^2 + 8x + 15)/(x-4)] * [(x^2 – 16) / 2x + 6)]
Answers
Answered by
Steve
x^2+8x+15 = (x+3)(x+5)
x^2-16 = (x+4)(x-4)
so, what you have is
(x+3)(x+5) / (x-4) * (x-4)(x+4) / 2(x+3)
Now you see that x+3 and x-4 cancel out, leaving
(x+5)(x-4)/2
x^2-16 = (x+4)(x-4)
so, what you have is
(x+3)(x+5) / (x-4) * (x-4)(x+4) / 2(x+3)
Now you see that x+3 and x-4 cancel out, leaving
(x+5)(x-4)/2
Answered by
Reiny
so ..
[(x^2 + 8x + 15)/(x-4)] * [(x^2 – 16) / (2x + 6)]
= (x+3)(x+5)/(x-4) * (x-4)(x+4)/(2(x+3) )
= (x+5)(x+4)/2 , x ≠ -3, 4
the temptation is to anticipate asymptotes at x = -3 and x = 4
but since we obtain 0/0 for these two values in the original, but an actual value for the final simplified form, we get a hole for these two values of x
Thus the restriction must be part of the answer.
[(x^2 + 8x + 15)/(x-4)] * [(x^2 – 16) / (2x + 6)]
= (x+3)(x+5)/(x-4) * (x-4)(x+4)/(2(x+3) )
= (x+5)(x+4)/2 , x ≠ -3, 4
the temptation is to anticipate asymptotes at x = -3 and x = 4
but since we obtain 0/0 for these two values in the original, but an actual value for the final simplified form, we get a hole for these two values of x
Thus the restriction must be part of the answer.
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