Yes. Consider the trinomial x2 + 2x + 4. It can be factored as (x+2)(x+2), that is to say, it has two identical factors of (x+2).
[ 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 -
It factors to: (x-4)(x+1)
That trinomial is unfactorable (the roots are not integers).
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