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[tex] {y}^{2} + 16y + 64 = (y + {?})(y + {?}[/tex]


Sagot :

Hi ^^

To find the missing terms, we only have to divide the "b" in the given by 2.

[tex]b = 16[/tex]

[tex]16 \div 2 = 8[/tex]

Then, we can check the answer by using the FOIL method.

(y + 8)(y + 8) = (y + 8)²

F:  (y)(y) = y²

O: (y)(8) = 8y

I:   (8)(y) = 8y

L:  (8)(8) = 64

= y² + 8y + 8y + 64

= y² + 16y + 64

The missing term is 8.

Answer:

y² + 16y = ( y + ) ( y + 8)

Explaination:

The square root of 64 is 8 and the square root of

y² is y

The middle term should be of the form 2ab. In this case, a=y and b= 8

So, 2(y)(8), or 16y, which is the middle term. This short proof justifies that it is indeed a perfect square trinomial.

y² - 16y + 64

= (y - 8)(y - 8)

= ( y - 8 )²

( y - 8 )²

hope it helps