x3 - 7x2 + 6x =x (x2 - 7x + 6) =x (x - 6) (x - 1)
7x2 plus 7x5 = 49
7x2 + 68xy - 20y2
x3 - 7x2 + 6x = x(x2 - 7x + 6) = x(x2 - x - 6x + 6) = x[x(x - 1) - 6(x - 1)] = x2(x - 1) - 6x(x - 1) = (x2 - 6x)(x - 1)
31
The term with the highest power(s) of the unknown variable(s) is 7x2. The power is 2 so the expression is a binomial.
x3 - 7x2 + 6x =x (x2 - 7x + 6) =x (x - 6) (x - 1)
With suitable signs between the terms, yes.
7x2 plus 7x5 = 49
7x2 + 68xy - 20y2
(7x2 + 9)(4x - 1)
x3 - 7x2 + 6x = x(x2 - 7x + 6) = x(x2 - x - 6x + 6) = x[x(x - 1) - 6(x - 1)] = x2(x - 1) - 6x(x - 1) = (x2 - 6x)(x - 1)
31
22 + y
Algebraic expression.
That equals 28
It is a quadratic polynomial.