How to factor (5r-6)(r+3)-(2r-1)(r+3)?

Answer 1

Hi Dear Kelly ;

(5r-6)(r+3)-(2r-1)(r+3) It is what we're ganna solve;

Step 1
Just think "FOIL" (first, outer, inner, last)
So
(5r-6)(r+3)
▪ Now,1. First! So multiply the first ones in the parentheses
So 5r *r = 5r^2
▪ Next, OUTER! So the outer numbers
5r * 3 = 15r
▪ Next, INNER! So do -6 * r = -6r
▪ Finally, LAST! So do -6 *3 = -18
Now add all the numbers up and
So you get ;
5r^2 + 15r - 6r - 18 = 5r^2 + [ 15 r - 6r ] - 18 = 5r^2 + 9r - 18

Step 2
do the same like what you did in the first step;
(2r-1)(r+3)
( 2r * r) + ( 2r * 3) + ( -1 * r) + ( -1 * 3 ) =
2r^2 + 6r - r - 3 = 2r^2 + 5r - 3

Step 3
Subtract the total results in step 1 and step 2
5r^2 + 9r - 18 - ( 2r^2 + 5r - 3 ) =
5r^2 + 9r - 18 - 2r^2 - 5r + 3 =
[ 5r^2 - 2r^2 ] + [ 9r - 5r ] + [ -18 + 3] =
3r^2 + 4r - 15

Good Luck Darling....

Answer 2

Since both sides are multiplied by (r+3), combine the terms:
(5r - 6)(r + 3) - (2r - 1)(r + 3) = (5r - 6 -2r + 1)(r + 3)
= (3r - 5)(r + 3)

Answer 3

Hi,

I think you mean to say to factorise
(5r-6)(r+3)-(2r-1)(r+3) = 0

here it is
(5r-6)(r+3)-(2r-1)(r+3) = 0
=> r = -3 or (5r-6 - 2r + 1) = 0
=> 3r = 7
=> r = 7/3 or -3

They are the factors.

Peace out.

Answer 4

Showing all the steps for the user's sake...
(5r-6)(r+3)-(2r-1)(r+3) =
(r+3)*((5r-6)-(2r-1)) =
(r+3)*(3r-5) =

Answer 5

Factor a (r+3) out:

(r+3)[(5r-6)-(2r-1)]

Simplify what's in the square brackets:

(r+3)(3r+7)

Answer 6

Split it into two parts -

(5r-6)(r+3)

(2r-1)(r+3)

Using the LIOR method would be easiest:

Left-hand pair
Inside pair
Outside Pair
Right-hand pair

Leaving:

5r²-6r+15r-18 = 5r²+9r-18

2r²-r+6r-3=2r²+5r-3


Now simplify:

5r²+9r-18-2r²+5r-3
= 3r²+14r-21

Answer 7

(5r-6)(r+3)-(2r-1)(r+3)
(r+3)(5r-6-2r+1)
(r+3)(3r-5)

Answer 8

I was taught in middle school to solve these sorts of mathematical problems in a specific order:Please Excuse My Dear Aunt Sally which means first parenthesis,exponents,multiplication,division,addition,then subtraction.