(r^2+7r+10)/3 * (3r-30)/(r^2-5r-50)
How can i Factorize each expression
to end up with (r+5)(r+2)/3 * 3(r-10)/((r-10)(r+5)
can you explain.
(r+5)(r+2)(3)(r-10)/[ 3(r-10)(r+5)]
r+5 cancels
3 cancels
r-10 cancels
so
= (r+2)
I know that they cancel but what i want to know its how did this expression
(r^2+7r+10)/3 * (3r-30)/(r^2-5r-50)
became this (r+5)(r+2)/3 * 3(r-10)/((r-10)(r+5)
like the first expression had a 7r where is that in the second.
The 7r term disappears in the factoring process.
The second order polynomials in the numerator and denominator were both factored by Damon. You may need to review the factoring process; it is an important part of algebra. You can multiply the two monomial factors back together to verify that
(r-10)(r+5) = r^2 -5r -50,
for example.
To factorize the expressions, we need to break them down into their highest common factors.
Let's start with the first expression: r^2 + 7r + 10.
We need to find two numbers that multiply to give 10 and add up to give 7 (the coefficient of the middle term, 7r). The numbers that satisfy this condition are 2 and 5. So, we can write the expression as:
(r + 2)(r + 5)
Now let's move on to the second expression: r^2 - 5r - 50.
Again, we need to find two numbers that multiply to give -50 and add up to give -5 (the coefficient of the middle term, -5r). The numbers that satisfy this condition are -10 and 5. So, we can write the expression as:
(r - 10)(r + 5)
Now that we have factored both expressions, we can rewrite the fraction:
(r^2 + 7r + 10)/3 * (3r - 30)/(r^2 - 5r - 50)
= [(r + 2)(r + 5)]/3 * [3(r - 10)]/[(r - 10)(r + 5)]
Now, since we have (r - 10) on both the numerator and the denominator, we can cancel them out:
= (r + 2)/3 * 3/(r + 5)
= (r + 2)(r + 5)/3 * 3/(r + 5)
Finally, we can simplify further:
= (r + 2)(r + 5)/3
So, the final factored form is (r + 2)(r + 5)/3.