If that's + 49y2, the answer is (9x - 7y)(9x - 7y) or (9x - 7y)2
(4x + 7y)(16x2 - 28xy + 49y2)
a proper factor is a factor.
a factor bug is a factor
A composite factor is a factor that is a composite number, as opposed to a prime factor which is a factor that is a prime number.
If that's + 49y2, the answer is (9x - 7y)(9x - 7y) or (9x - 7y)2
This is a difference of two sqaures. 49y2 - z2 = (7y - z)(7y + z)
No.
16x2+56xy+49y2
(7y-12)(7y+12)
4x^(2) + 28xy + 49y^(2) -> (2x)^(2) + 28xy + (7y)^2 Split into brackets ( 2x + 7y)(2x + 7y) If you applyuFOIL you will find toy come to '28xy'. The clue to this problem is to note that '4' ^ '49' are both squared numbers.
(4x + 7y)(16x2 - 28xy + 49y2)
Do you mean some thing like 4x2 - 28xy + 49y2 ? notice 4x2 = (2x)2 and 49y2 = either (7y)2 or (-7y)2 (1st and last terms must be squares) and the middle term is 2(2x)(-7y) (2 times sq root of 1st times sq root of last) Once the pattern is recognized: (sq root 1st + sqroot 2nd)2 or (sq root 1st - sqroot 2nd)2 depending on if middle term is positive or negative. So, back to my example: 4x2 - 28xy + 49y2 = (2x - 7y)2 because middle is negative. Simpler example factor x2 + 6x + 9. 1st and last are both squares sq root of 9 = 3 3 times 2 = 6 that fits the middle so x2 + 6x + 9 = (x + 3)2
Factors are (7y - 3)(7y - 2) so it's not a perfect square.
25x2-49y2 is the difference of two squares and can be factored as:- (5x-7y)(5x+7y)
125x3+343y3 can also be written (5x)3+(7y)3. It helps to rewrite the constants like this because there is a general factoring rule: a3+b3 = (a+b)(a2-ab+b2). In this example, the a is 5x and the b is 7y. Plug in and get (5x+7y)(25x2-35xy+49y2).
In order to factor the sum of the cubes, we need to use this form a³ + b³ = (a + b)(a² - ab + b²). Let a³ = 27x³ and b³ = 343y³. Then, a = 3x and b = 7y. Perform substitution of these terms for the form, and we obtain: 27x3 + 343y3 = (3x + 7y)(9x2 - 21xy + 49y2)