[ x3 + 3x2 + 2x ] is a trinomial. It's factors are [ x, (x + 1), (x + 2) ] .
The sum of p and q
The sum of -p and -q -
To find the factors of the trinomial (3m^2 + 11mn + 6n^2), we need to break it down into two binomials. First, we find two numbers that multiply to the product of the leading coefficient and constant term, which are (3 \times 6 = 18). Then, we look for two numbers that add up to the middle coefficient, which is 11. The factors are ((3m + 2n)(m + 3n)).
It factors to: (x-4)(x+1)
It is 1 if the two are the only factors.
Each has two binomial factors.
-- If the last term of the trinomial ... the one that's just a number with no 'x' ... is positive, then both factors have the same sign as the middle term of the trinomial. -- If the last term is negative, then the factors have different signs. If this was never pointed out in class, well, it should have been.
10 and 10
Homozygous
A square number.
The sum of -p and -q -
The sum of -p and -q -
Five identical prime factors : 32, 96. Four identical prime factors : 16, 48, 80, as well as 81. Three identical prime factors : 8, 24, 40, 56, 72, 88, as well as 27, 54. Two identical prime factors : All multiples of 4 not yet listed (4, 12, 20...), all multiples of 9 not yet listed (9, 18, 36...), as well as 25, 50, 75, 100, and 49 and 98. Your teacher forgot "six identical prime factors" : 64.
[ x3 + 3x2 + 2x ] is a trinomial. It's factors are [ x, (x + 1), (x + 2) ] .
The sum of p and q
Yes; the factored form would be (9c+4)(9c+4) or just (9c+4)2 Since the two factors are the same, the beginning trinomial 81c2+72c+16 is a perfect square trinomial