factor: 21x^3-35xy^2?

2 ⤊

2006-12-30 06:19:16 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

Both terms have a 7x in common. When you factor that out, you are left with:

7x(3x^2-5y^2).

2006-12-30 06:27:34 · answer #1 · answered by wizard of ozma 3 · 0⤊ 0⤋

21x^3-35xy^2
7x(3x^2-5y^2)
7x(√3 x-√5 y)(√3 x+√5 y)

2006-12-30 06:22:11 · answer #2 · answered by yupchagee 7 · 1⤊ 0⤋

First: find a number divisible by the coefficients which is 7 >

= 7(3x^3 - 5xy^2)

2006-12-30 06:23:48 · answer #3 · answered by ♪♥Annie♥♪ 6 · 0⤊ 1⤋

Take out common factor 7x:

7x(3x^2 - 5y^2)

2006-12-30 06:21:57 · answer #4 · answered by martina_ie 3 · 1⤊ 1⤋

21x³ - 35xy²

7x(3x² - 5y²)

- - - - - - s-

2006-12-30 07:46:01 · answer #5 · answered by SAMUEL D 7 · 0⤊ 0⤋