How do I factor this using " special identities" ?

Question

Hi , I need help understanding how to solve this. Can you please explain with steps?  The question is to factor :  16y^2 - 56y + 49 Using the "special identity " :   a^2 +2ab + b^2. 

Answer 1

The "Special Identity"  is:

        a² +2ab + b²    =    (     a  +     b  )²         [verify using FOIL)

 

The problem is:

      16y²   -      56y     +   49           so    a=4y     and    b = -7

        a²    +     2ab      +   b²

      (4y)²  + 2(4y)(-7) + (-7)²    =   (  (4y)   +  (-7)  )²   [see "Special Identity" above]

        factored as:    ( 4y - 7 )²

Answer 2

a^2+2ab+b^2 is a perfect square

16y^2-56y+49 is a perfect square

(4y)^2=16y^2

7^2=49

a=4y, b=-7, 2ab=2*(4y)*(-7)=-56y

the polynomial factors into...

(4y-7)(4y-7)=16y^2-28y-28y+49=16y^2-56y+49

Answer 3

Alex,

a² + 2ab + b² is referred to as a perfect square and always factors as (a + b)².  But your problem says -56y, so your identity is a² - 2ab + b^2. . Still a perfect square, but factors as (a - b)².

 

So for this problem, a² = 16  ∴  a = 4

and b² = 49  ∴  b = 7

so 2ab = 2(4)(7) = 56, verification of form.

 

so the factored form is:   (4y - 7)²