If you mean: x2+3x+2 then it is (x+1)(x+2) when factored
3x2 - 5x - 2 can be factored into (3x + 1) (x - 2)
If you mean: (x+2) and (x+1) then it is x^2+3x+2
(2x + 5)/(3x + 2); √(x² - 2x + 3) a
To find the roots of the polynomial (x^2 + 3x - 5), we need to set the polynomial equal to zero and solve for x. So, (x^2 + 3x - 5 = 0). To solve this quadratic equation, we can use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where a = 1, b = 3, and c = -5. Plugging these values into the formula, we get (x = \frac{-3 \pm \sqrt{3^2 - 41(-5)}}{2*1}), which simplifies to (x = \frac{-3 \pm \sqrt{29}}{2}). Therefore, the two values of x that are roots of the polynomial are (x = \frac{-3 + \sqrt{29}}{2}) and (x = \frac{-3 - \sqrt{29}}{2}).
If you mean: x2+3x+2 then it is (x+1)(x+2) when factored
(x + 1)(x + 2)
Quotient =3x 3 −x 2 −x−4 Remainder =−5
3x2 + 2x - 8 = 3x2 + 6x - 4x - 8 = 3x(x+2) - 4(x+2) = (3x-4)*(x+2)
15x^2+3x-12 3(5x^2+x-4)=Answer
(x + 1) and (x + 2) are monomial factors of the polynomial x2 + 3x + 2 (x + 1) and (x + 3) are monomial factors of the polynomial x2 + 4x + 3 (x + 1) is a common monomial factor of the polynomials x2 + 3x + 2 and x2 + 4x + 3
3x2-9x-30 =3x2+6x-15x-30 =3x(x+2)-15(x+2) =(3x-15)(x+2) OR 3(x - 5)(x+2)
5x2 + 3x - 2 is a polynomial so, we can say: p(x) = 5x2 + 3x - 2which is of the form of ax2 + bx + cFirst of all multiply the coefficient of x2 with cHere a = 5 and c = -2ac = 5(-2) = -10Break the coefficient of x i.e. b into two numbers such that the product of the numbers is equal to ac.We can't break 3 into two numbers so that their product is -10.So, it is not possible to factor p(x).
The product of (3x-1)(x+5) = 3x^2 +14x -5 by multiply out the brackets
To divide (x^2 + 3x - 2) by (x - 2), you can use polynomial long division or synthetic division. The result of dividing these two polynomials is (x + 5), with a remainder of 8.
x2 + 3x - 18 = (x + 6)(x - 3)
3x