64y^3 + 80y^2 + 25y
a. y(8y + 5)(8y - 5)
b. y(64y + 1)(y + 25)
c. y(8y + 5)^2
d. y(8y - 5)^2
factor the polynomial?
2009-05-29 4:08 pm
回答 (7)
2009-05-29 4:13 pm
✔ 最佳答案
64y³ + 80y² + 25y =y( 64y² + 80y + 25) =
y( 8y + 5)²
2009-05-29 5:39 pm
64y^3 + 80y^2 + 25y
= y(64y^2 + 80y + 25)
= y(64y^2 + 40y + 40y + 25)
= y[(64y^2 + 40y) + (40y + 25)]
= y[8y(8y + 5)) + 5(8y + 5)]
= y(8y + 5)(8y + 5)
= y(8y + 5)^2
(answer c)
= y(64y^2 + 80y + 25)
= y(64y^2 + 40y + 40y + 25)
= y[(64y^2 + 40y) + (40y + 25)]
= y[8y(8y + 5)) + 5(8y + 5)]
= y(8y + 5)(8y + 5)
= y(8y + 5)^2
(answer c)
2009-05-29 4:35 pm
c
2009-05-29 4:35 pm
you don't really have to factor this during a test
use the process of elimination, which is easy here
first, factor the y out, and we're left with 64y² + 80y + 25
a. NOT! because 5 * (-5) = - 25 but we need + 25
b. NOT! because 64y * 25 will give us a huge number of y terms
c. YES! 8y * 8y gives us 64y² 40y + 40y gives us 80y and 5 * 5 gives us the 25 term
next
use the process of elimination, which is easy here
first, factor the y out, and we're left with 64y² + 80y + 25
a. NOT! because 5 * (-5) = - 25 but we need + 25
b. NOT! because 64y * 25 will give us a huge number of y terms
c. YES! 8y * 8y gives us 64y² 40y + 40y gives us 80y and 5 * 5 gives us the 25 term
next
2009-05-29 4:30 pm
y(8y+5)^2
c is correct.
c is correct.
2009-05-29 4:14 pm
y(64y² + 80y + 25)
y(8y + 5)(8y + 5)
y(8y + 5)²
y(8y + 5)(8y + 5)
y(8y + 5)²
2009-05-29 4:14 pm
64y^3 + 80y^2 + 25y
y(64y^2 + 80y + 25)
y(8y^2 + 5)^2
Answer is C.
y(64y^2 + 80y + 25)
y(8y^2 + 5)^2
Answer is C.
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