x2 + 36 cannot be factored. You can only factor the difference of two squares, not the sum.
(x + 2)(x + 1)
x2 + 8xy + 15y2 = (x + 3y) (x + 5y)
3(x + 4)(x + 1)
(x-1)(x-4)
(x - 5)(x + 12)
(x + 12)(x - 3)
2x^(2) + 72 Factors to 2( x^(2) + 36) s(x^(2) + 6^(2)) Does NOT factor. NB Remember . two squared terms with a positive(+) between them does NOT factor!!!! However, two squared terms with a negative(-) between DOES factor . e.g. x^(2) + 6^(2) Does NOT factor x^(2) - 6^(2) factors to ( x - 6)( x + 6 ) Note the different signs. Similarly 8^(2) + 6^(2) does NOT factor 8^(2) - 6^(2) factors to (8 - 6)(8 + 6) Or using the ~Pythagorean Equation. h^(2) = a^(2) + b^(2) Does NOT factors However, a^(2) = h^(2) - b^(2) factors to a^(2) = (h - b)(h + b) .
(6x - 1)(6x - 1)
3x2-21x+36 =3(x2-7x+12) =3(x-4)(x-3)
9x2 + 27x - 36 = 9(x2 + 3x - 4) = 9(x + 4)(x - 1)
(a + x^2)(b + y^2)
No
Since the problem has 4 terms, first you factor x cubed plus 9x squared, then you factor 2x plus 18. So when you factor the first two term, you would get x sqaured (x plus 9). Then when you factor the last two terms and you get 2 (x plus 9). Ypure final answer would be (x squared plus 2)(x plus 9)
x2y + axy + abx + a2b Factor by grouping. xy(x + a) + ab(x + a) (xy + ab)(x + a)
(x + 6)(x - 6)
(x+1)(x+8)
(x + 2)(x + 1)