(2p - 3)(2p + 1)
Two. 2 and p Since p is prime it's only factor (other than one) is it's self.
It has four factors: 1, 2, p, and 2p.
0.44444
If a fraction is equivalent to one-half then the numerator is always half the denominator. Or, the double the numerator is the denominator. So, given any other fraction, p/q, if 2p < q then the fraction < 1/2 if 2p = q then the fraction = 1/2 if 2p > q then the fraction > 1/2
The GCF is 2p.
The greatest common factor of 5p and 20pp is 5p.Factors of 20pp can be written out as 1p and 20p, 2p and 10p, 4p and 5p.Factors of 5p can be written as 1 and 5p or 1p and 5.The highest factor they both have in common is 5p.
(2p - 3)(2p + 1)
p2-2p = p(p-2) when factored
2p + 11p + 5 = 13p + 5 which does not factor. If the expression had been 2p2 + 11p + 5 then this can be factored, 2p2 + 11p + 5 = (2p + 1)(p + 5)
3p2-2p-5 = (3p-5)(p+1) when factored
If we have 6p+3 The common factor to both 6p and 3 is 3. If we take this out the front of brackets we get 3(2p+1)
2(2p + 5q)
This is an expression, not an equation. To rearrange it, you arrange the expression so that the first set of terms have a variable (letter) in common. Arrange the rest so that they follow a similar patter (see working below). Then factorise. 2pr + 6qs - 3qr - 4ps. Select r to be the common variable for the first two terms. So 2pr - 3qr + 6qs - 4ps You will see that terms (now numbered) 3 and 4 have s in common. But, while with the first two, the term with p comes before the term with q, with 3 and 4, it is the other way around. So swap terms 3 and 4. 2pr - 3qr - 4ps + 6qs Common factor of terms 1 and 2 = r, common factor of terms 3 and 4 = s: r(2p - 3q) - 2s(2p - 3q) and then (2p - 3q) is a common factor so (2p - 3q)(r - 2s)
If all the terms are positive then it is: (2p+3)(p+1)
(2p - 5)(5p^2 + 2)
True Observe the following: 5x + 5y + 3z = 5(x + y) + 3z The first two terms could be factored because they shared a common factor of 5, but the third term did not -- not all terms need to share a common factor to use the grouping method.