We can't answer that without some polynomials to choose from.
(3x + 4)(3x + 4)
3x^2 + 4x, among many others.
(3x + 4)(2x - 1)
15x^2+3x-12 3(5x^2+x-4)=Answer
4
(3x + 4)(3x + 4)
3x^2 + 4x, among many others.
(6x-4) = 2 (3x-2)
(3x + 4)(2x - 1)
15x^2+3x-12 3(5x^2+x-4)=Answer
(x-4) (x+7)
7
4
To find the roots of the polynomial (3x^5 + 2x^3 + 3x), we can factor out the common term, which is (x): [ x(3x^4 + 2x^2 + 3) = 0. ] This shows that (x = 0) is one root. The quartic polynomial (3x^4 + 2x^2 + 3) does not have real roots (as its discriminant is negative), meaning it contributes no additional real roots. Therefore, the polynomial has only one real root, which is (x = 0).
2y(3x + 1)
(3x + 1)(x - 5)
3x2 + 2x - 8 = 3x2 + 6x - 4x - 8 = 3x(x+2) - 4(x+2) = (3x-4)*(x+2)