factor completely?

Question

1.

22y^4-33y^3+11y^2




2.

77r^7s^7-84r^8s^4

Answer 1

OK

22y^4-33y^3+11y^2 =
11y^2(2y^2 - 3y +1)=

11y^2(2y -1)(y-1)

===============

77r^7s^7-84r^8s^4=

7r^7s^4(11s^3 - 12r)

============

Hope that helps.

Answer 2

1) 22y^4-33y^3+11y^2

Simplifying
22y^4 + -33y^3 + 11y^2

Reorder the terms:
11y^2 + -33y^3 + 22y^4

Factor out the Greatest Common Factor (GCF), '11y^2'.
11y^2(1 + -3y + 2y^2)

Factor a trinomial.
11y62((1 + -2y)(1 + -1y))

Final result:
11y^2(1 + -2y)(1 + -1y)




2) 77r^7s^7-84r^8s^4

Simplifying
77r^7s^7 + -84r^8s^4

Factor out the Greatest Common Factor (GCF), '7r^7s^4'.
7r^7s^4(11s^3 + -12r)

Final result:
7r^7s^4(11s^3 + -12r)

Answer 3

22y^4 - 33y^3 + 11y^2

11y^2(2y^2 - 3y + 1) {common factor}

11y^2(2y - 1)(y - 1)


77r^7s^7 - 84r^8s^4

7r^7s^4(11 s^3 - 12r)

Answer 4

22y^4-33y^3+11y^2
11y^2(2y^2-3y+1) = 11y^2(2y-1)(y-1)

77r^7s^7-84r^8s^4
7r^7s^4(11s^3 -12r)

Answer 5

11y²(2y² - 3y + 1)
11y²(2y - 1)(y - 1)

7r^7s^4(11s^3 - 12r)

Answer 6

#1
11y^2 (2y^2 - 3y +1)
11y^2 (2y - 1) (x - 1)


#2
7r^7s^4 (11s^3 - 12r)

Answer 7

1. 11y^2(2y-1)(y-1)

2. 7r^7s^4(11s^3-12r)