What are the factors of 5y^2-2y-7

7/5, -1

Step-by-step explanation:

all quadratic equations have 2 roots so meaning it is factorized into to bracts so first step

(5y      )(y      )

the reason there is a five before one of the y's is because the a in our function {ax^2+bx+c}

second step is to factorize so u think what 2 numbers are equal to -7 when multiplied. since 7 is a prime number its either 7 and -1 or -7 and 1

so u try both factors to check the middle term

first  one

(5y+7)(y-1)

that would be equal to 5y^2+2y-7 which isn't the function we are looking for therefore the

second one (5y-7)(y+1)

that would be equal to 5y^2-2y-7 which is correct

it would be:

c.parallelogram.