(x - 2)(x + 2)(x - 3)(x + 3)
x2 + 36 cannot be factored. You can only factor the difference of two squares, not the sum.
(5w-6)2
4 is a factor of 36 and 36 is a multiple of 4
2x + 6 = 2(x + 3) so 2x + 6 = 0 implies x = -3 Now, substitute x = -3 in 8x3 + 20x2 + 36 which is -8*27 + 20*9 + 36 = -216 + 180 + 36 = 0 So -3 is a root of 8x3 + 20x2 + 36 ie, by the remainder theorem, (x + 3) is a factor. Also, each of the coeffivcients in the expression is even, so 2 is also a factor. So 2*(x + 3) or (2x + 6) is a factor.
No. 36 is composite and so it cannot be a prime factor.
x(x2 + 36)
x^(2) + 13x + 36 Factors to ( x + 9)(x + 4) When learning factoring. ;- #1 ; If the coefficient of x^(2) is '1' , as in this case. Then #2 ; Write down all the factors of 36, which are ,1,36 ' 2,18 ' 3,12 ' 4,9 ' 6,6 ; #3 ; Out of these pairs of factors, select a pair that add/subtract to '13'. #5, They are 4,9 ; 4 + 9 = 13 #6 ; Since the quadratic eq;m has positive (+) signs , then all the signs in the brackets are positive(+). When the coefficient of x^(2) is > 1, and/or the signs are different , then different techniques come into play.
x2 + 36 cannot be factored. You can only factor the difference of two squares, not the sum.
(x + 4)(x + 9)
(x + 9)(x + 4)
2(x + 6)(x + 3)
x^(2) - 13x + 36 Factors to ( x - 9)(x - 4)
6x3 - 27x2 + 3x + 36 = 3 (x - 4) (x + 1) (2x - 3)
(5w-6)2
(6 - 5x)(6 - 5x)
(x - 6)(5x - 6)
(2x - 3)(x - 12)