How do you factor $16 y^{2} - 25$ $16y^2-25$ ?

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How do you factor $16 y^{2} - 25$ $16y^2-25$ ?

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Answer 1 · 2018-03-21T00:41:44

$(4 y + 5) (4 y - 5)$ $(4y+5)(4y-5)$
You need to consider what multiplies to make 16(either $1 \cdot 16, 2 \cdot 8,$ $1*16, 2*8,$ or $4 \cdot 4$ $4*4$ ), and what multiplies to make 25( $5 \cdot 5$ $5*5$ ). Also notice that this is a binomial, not a trinomial.

Explanation:

The only factor of $25$ $25$ is $5 \cdot 5 = 5^{2}$ $5*5=5^2$ , so the factorization must be of the form $(a + 5) (b - 5)$ $(a+5)(b-5)$ , as a negative times a positive is negative. Now consider that there is no middle term, so it must have been canceled out. This implies that the coefficients of $y$ $y$ are the same. This leaves only $(4 y + 5) (4 y - 5)$ $(4y+5)(4y-5)$ .

Answer 2 · 2018-03-21T00:42:42

As the difference between two squares.

Explanation:

$16 y^{2} - 25 = (4 y - 5) \times (4 y + 5)$ $16y^2 - 25 = ( 4y -5) xx ( 4y +5)$

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How do you factor $16 y^{2} - 25$ $16y^2-25$ ?

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