13y^2 - 5yx - 8x^2 Factor completely
To factor the polynomial 13y^2 - 5yx - 8x^2 completely, we look for two binomials that multiply together to give us the original polynomial.
The first terms of these binomials should be factors of 13y^2, and the last terms should be factors of -8x^2. The middle terms should be the product of the outer and inner terms.
The factors of 13y^2 are either 1y and 13y or 13y and 1y.
The factors of -8x^2 are either 1x and -8x or -8x and 1x.
Now, we try different combinations to see which one satisfies the condition for the middle terms.
Combination 1: (1y - 8x)(13y + 1x)
= 13y^2 - 64x^2 + 1xy - 8xy
= 13y^2 -72x^2 - 7xy
-7xy is not the same as -5yx, so this combination does not work.
Combination 2: (1y + 8x)(13y - 1x)
= 13y^2 - 1x^2 + 104xy - 1xy
= 13y^2 -9x^2 + 103xy
103xy is not the same as -5yx, so this combination does not work.
Combination 3: (13y - 8x)(1y + x)
= 13y^2 + 13xy - 8xy - 8x^2
= 13y^2 + 5xy - 8x^2
This combination satisfies the conditions for the middle terms.
Therefore, the polynomial 13y^2 - 5yx - 8x^2 can be factored completely as (13y - 8x)(y + x).