gcf(35x², 7xy², 21x²y²) = 7x
35x² = 5 times 7 times x²
7xy² = 7 times x times y²
21x²y² = 3 times 7 times x² times y²
hcf = 7 times x = 7x
(I've written the multiplication as times instead of using the × symbol to avoid confusion with the unknown x.)
3x+21x+9x=33x. Break it apart. The factor of x is x and the factors of 33 are 3 and 11. Therefore, the factors of 3x+21x+9x are 3, 11, and x.
3(7x + 11)
7x
21x^2
i hope you mean (2x^2 + 21x + 49) you cant solve it its already in its lowest possible form
The greatest common factor (GCF) is 7.
x
To factor the expression 21x^35, you need to find the common factors of 21 and x^35. The factors of 21 are 1, 3, 7, and 21. The factors of x^35 are x, x^2, x^3, ..., x^35. Therefore, the factored form of 21x^35 is 21x*x^34 or 21x^(35-1).
14xy - 21x + 7 '7' is a common factor to all three coefficients. So 'take it out'. Hence 7(2xy - 3x + 1) The rest of it , inside the brackets does not factor. However, I suspect the 'y' should be a 'power of 2' Hence 7(2x^(2) - 3x + 1) This factors to 7(2x - 1)(x - 1) Fully factored.!!!!
3x+21x+9x=33x. Break it apart. The factor of x is x and the factors of 33 are 3 and 11. Therefore, the factors of 3x+21x+9x are 3, 11, and x.
Make it equal zero by subtracting 21x from both sides. 20x2 - 21x - 5 factors to (4x - 5)(5x + 1)
3(7x-6)
3(7x + 11)
It is: 3x*(x+7)
(x - 8)(x - 13)
(2x - 1)(x - 10)
3x2-21x+36 =3(x2-7x+12) =3(x-4)(x-3)