Question

What is the product of 2r^2-5 and 3r

Answer 1

Answer: Hence, our required product would be

$6r^3-15r$

Step-by-step explanation:

Since we have given that

$(2r^2-5)\ and\ 3r$

We need to find the product.

So, the product would be

$3r(2r^2-5)\\\\=3r\times 2r^2-3r\times 5\\\\=6r^3-15r$

Hence, our required product would be

$6r^3-15r$

Answer 2

Two solutions were found :

r = -5/2 = -2.500

r = 1

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :  

                    2*r^2-(5-3*r)=0  

Step by step solution :

Step  1  :

Equation at the end of step  1  :

 2r2 -  (5 - 3r)  = 0  

Step  2  :

Trying to factor by splitting the middle term

2.1     Factoring  2r2+3r-5  

The first term is,  2r2  its coefficient is  2 .

The middle term is,  +3r  its coefficient is  3 .

The last term, "the constant", is  -5  

Step-1 : Multiply the coefficient of the first term by the constant   2 • -5 = -10  

Step-2 : Find two factors of  -10  whose sum equals the coefficient of the middle term, which is   3 .

     -10    +    1    =    -9  

     -5    +    2    =    -3  

     -2    +    5    =    3    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -2  and  5  

                    2r2 - 2r + 5r - 5

Step-4 : Add up the first 2 terms, pulling out like factors :

                   2r • (r-1)

             Add up the last 2 terms, pulling out common factors :

                   5 • (r-1)

Step-5 : Add up the four terms of step 4 :

                   (2r+5)  •  (r-1)

            Which is the desired factorization