Simplify the expression [(x^2-6x+5)/(x+3)]*[(x^2-9)/(x-5).
- print Print
- list Cite
Expert Answers
calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
Let E = (x^2-6x+5) / (x+3) * (x^2-9)/(x-5)
First we will multiply numerators and denominators:
==> E = (x^2-6x+5)*(x^2-9)/(x+3)(x-5)
Now we will factor the numerator:
==> E = (x-5)(x-1)(x-3)(x+3)/(x+3)(x-5)
Now we will reduce similar terms.
==> E = (x-1)(x-3)
==> E = (x-1)(x-3) = x^2 -4x + 3
Related Questions
- How we simplify 2x/(x^2 -6x+9) - 1/(x+1) - 8/(x^2 -2x-3)
- 2 Educator Answers
- Simplify (x-2)^2
- 1 Educator Answer
- 2 x 3^x = 7 x 5^x
- 2 Educator Answers
- `f(x) = x^3 - 6x^2 + 5, [-3, 5]` Find the absolute maximum and minimum values of f on the...
- 3 Educator Answers
- Simplify the expression 2(a -3) + 4b - 2(a -b -3) + 5
- 1 Educator Answer
To simplify {(x^2-6x+5)/(x+3)}{(x^2-9)/(x+5)}.
Solution:
x^2--6x+5 = (x-1)(x-5)
x^2-9 = (x-3)(x+3)
So the given expression is rewritten sing the above factors:
{(x-1)(x-5)/(x+3)}/{(x-3)(x+3)/(x-5)}
=(x-1)(x-5)(x-3)(x+3)/{(x+3)((x-5)
=(x-1)(x-3)
= x^2-4x+3.
Therefore [(x^2-6x+5)/(x+3)]*{(x^2-9)/(x-5)} = x^2-4x+3.
We'll factorize all fractions in distinct factors. For this reason, we'll determine the roots of the first numerator:
x^2-6x+5 = 0
We'll apply quadratic formula:
x1 = [6+sqrt(36 - 20)]/2
x1 = (6+4)/2
x1 = 5
x2 = (6-4)/2
x2 = 1
The equation will be written as:
x^2-6x+5= (x-5)(x-1)
We'll write the numerator of the 2nd factor as a product:
x^2 - 9 = (x-3)(x+3)
We'll re-write the factorised expression:
[(x-5)(x-1)/(x+3)]*[(x-3)(x+3)/(x-5)]
We'll cancel common factors:
(x-1)(x-3)
We'll leave the factors as they are and the simplified expression is:
(x-1)(x-3)
We'll remove the brackets:
(x-1)(x-3) = x^2 - x - 3x + 3
We'll combine like terms:
(x-1)(x-3) = x^2 - 4x + 3
The simplified expression is:
x^2 - 4x + 3
Student Answers