Question

y2 - 20y + 100

Factorize the following problem with suitable identity​

Answer 1

Answer:

hey mate,

Step-by-step explanation:

Step by Step Solution

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Reformatting the input :

Changes made to your input should not affect the solution:

(1): "y2"   was replaced by   "y^2".  

STEP1 :

Trying to factor by splitting the middle term

1.1     Factoring  y2-20y+100  

The first term is,  y2  its coefficient is  1 .

The middle term is,  -20y  its coefficient is  -20 .

The last term, "the constant", is  +100  

Step-1 : Multiply the coefficient of the first term by the constant   1 • 100 = 100  

Step-2 : Find two factors of  100  whose sum equals the coefficient of the middle term, which is   -20 .

     -100    +    -1    =    -101  

     -50    +    -2    =    -52  

     -25    +    -4    =    -29  

     -20    +    -5    =    -25  

     -10    +    -10    =    -20    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -10  and  -10  

                    y2 - 10y - 10y - 100

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

                   y • (y-10)

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

                   10 • (y-10)

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

                   (y-10)  •  (y-10)

            Which is the desired factorization

Multiplying Exponential Expressions:

1.2    Multiply  (y-10)  by  (y-10)  

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (y-10)  and the exponents are :

         1 , as  (y-10)  is the same number as  (y-10)1  

and   1 , as  (y-10)  is the same number as  (y-10)1  

The product is therefore,  (y-10)(1+1) = (y-10)2  

=   (y - 10)2 final answer

thank you