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.
(r^2+7r+10)/3 * (3r-30)/(r^2-5r-50)
how can i Factorize each expression to end up with this new equation
(r+5)(r+2)/3 * 3(r-10)/((r-10)(r+5)
can you please explain..
2 answers
Let's factor r^2 + 7r + 10.
When we factor a trinomial, we get the
product of 2 binomials.
1. First, we replace r^2 with r*r:
(r + )(r + ),
2. We eliminate 7r, and 10 by finding 2 numbers whose product equals 10 and
their sum equals 7:
10 = 1*10 = 2*5.
We select 2, and 5 because their sum equals 7 and their product equals 10.
The factored trinominal equals:
(r + 2) (r + 5)
This method of factoring is convient only when the coefficient of X^2 = 1.
CHECK: Multiply each term in the 1st
parenthesis by the quantity in the 2nd parenthesis:
r(r + 5) + 2(r + 5),
Do the Multiplication:
r^2 + 5r + 2r + 10,
Combine like-terms and get:
r^2 + 7r + 10.
You try factoring the 2nd trinomial.
HINT: -50 = 1(-50) = 2(-25) = 5(-10).
When we factor a trinomial, we get the
product of 2 binomials.
1. First, we replace r^2 with r*r:
(r + )(r + ),
2. We eliminate 7r, and 10 by finding 2 numbers whose product equals 10 and
their sum equals 7:
10 = 1*10 = 2*5.
We select 2, and 5 because their sum equals 7 and their product equals 10.
The factored trinominal equals:
(r + 2) (r + 5)
This method of factoring is convient only when the coefficient of X^2 = 1.
CHECK: Multiply each term in the 1st
parenthesis by the quantity in the 2nd parenthesis:
r(r + 5) + 2(r + 5),
Do the Multiplication:
r^2 + 5r + 2r + 10,
Combine like-terms and get:
r^2 + 7r + 10.
You try factoring the 2nd trinomial.
HINT: -50 = 1(-50) = 2(-25) = 5(-10).
1 answer