Factor 98xy^4-28xy^2z+2xz^2

$98 x y^{4} - 28 x y^{2} z + 2 x z^{2}$

Step 1

Factor $2 x$ out of $98 x y^{4} - 28 x y^{2} z + 2 x z^{2}$ .

Factor $2 x$ out of $98 x y^{4}$ .

$2 x (49 y^{4}) - 28 x y^{2} z + 2 x z^{2}$

Factor $2 x$ out of $- 28 x y^{2} z$ .

$2 x (49 y^{4}) + 2 x (- 14 y^{2} z) + 2 x z^{2}$

Factor $2 x$ out of $2 x z^{2}$ .

$2 x (49 y^{4}) + 2 x (- 14 y^{2} z) + 2 x (z^{2})$

Factor $2 x$ out of $2 x (49 y^{4}) + 2 x (- 14 y^{2} z)$ .

$2 x (49 y^{4} - 14 y^{2} z) + 2 x (z^{2})$

Factor $2 x$ out of $2 x (49 y^{4} - 14 y^{2} z) + 2 x (z^{2})$ .

$2 x (49 y^{4} - 14 y^{2} z + z^{2})$

Step 2

Factor using the perfect square rule.

Rewrite $49 y^{4}$ as ${(7 y^{2})}^{2}$ .

$2 x ({(7 y^{2})}^{2} - 14 y^{2} z + z^{2})$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$14 y^{2} z = 2 \cdot (7 y^{2}) \cdot z$

Rewrite the polynomial.

$2 x ({(7 y^{2})}^{2} - 2 \cdot (7 y^{2}) \cdot z + z^{2})$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 7 y^{2}$ and $b = z$ .

$2 x {(7 y^{2} - z)}^{2}$