 Solving quadratic equations by factoring, Vieta’s formula Quadratic function or the second-degree polynomial

Roots or zeros of the function, axis of symmetry and y-intercept
A quadratic equation is a polynomial equation of the second degree. The general or standard form is
ax2 + bx + c = 0,   a is not 0
where a, b and c are coefficients.
The solutions of the equation are called roots (which may be real or complex) and are given by the quadratic formula.
We use the completing the square method to derive the quadratic formula to find the roots, The roots can also be written If  c = 0, the roots of the equation         ax2 + bx = 0       we can obtain by factoring
x(ax + b) = 0
or by using the quadratic formula, so we get   x1= 0  and  x2 = -b/a.
Solving quadratic equations by factoring, Vieta’s formula
A quadratic trinomial  ax2 + bx + can be factorized as
 ax2 + bx + c = a·[x2 + (b/a)·x + c/a] = a·(x - x1)(x - x2), where x1 + x2 = b/a and  x1· x2 = c/a
That means, to factor a quadratic trinomial we should find such a pair of numbers x1 and x2 whose sum equals b/a and whose product equals c/a.
So, if a > 0 and the constant term c negative, then the signs of x1 and x2 will be different while, when c is positive, their signs will be the same.
Then we use zero product principle to solve the quadratic equation, that is, a product a · b = 0 if at least one of its factors is 0.
 Examples:     Find the roots of each quadratic equation by factoring: a)  x2 - 3x -10 = 0 since   x2 - 3x -10 = x2 + (-5 + 2)·x + (-5)·(+2) = x2 - 5x + 2x -10 = x · (x - 5) + 2 · (x - 5) = (x - 5) · (x + 2), then     x - 5 = 0    =>     x1 = 5,     or      x + 2 = 0    =>     x2 = -2. b)  2x2 - 7x + 3 = 0 as  2x2 - 7x + 3 = 2[x2 - (7/2)x + 3/2] =  2[x2 - (1/2)x - 3x + 3/2] = = 2[x(x - 1/2) - 3(x - 1/2)] = 2 (x - 1/2)(x - 3) = (2x - 1) (x - 3), then     2x - 1 = 0    =>     x1 = 1/2,     or      x - 3 = 0    =>     x2 = 3 c)  3x2 - x - 2 = 0 as  3x2 - x - 2 = 3[x2 - (1/3)x - 2/3] =  3[x2 + (2/3)x - x - 2/3] = = 3[x(x + 2/3) - (x + 2/3)] = 3·(x + 2/3)(x - 1) = (3x + 2)(x - 1), then     3x + 2 = 0    =>     x1 = -2/3,     or      x - 1 = 0    =>     x2 = 1.
Example:   Given are leading coefficient a2 = -1 and the pair of conjugate complex roots,
x1 = 1 + and  x2 = 1 - i, of a second degree polynomial;
a)  find the polynomial using the above theorem,
b)  make a check of the solutions of the polynomial using the quadratic formula.
 Solution: a)  By plugging the given values into    a2x2 + a1x + a0 = a2(x - x1) (x - x2) a2x2 + a1x + a0 = -1[x - (1 + i)] · [x - (1 - i)] = -[(x - 1) - i] · [(x - 1) + i] = = -[(x - 1)2 - i2] = -(x2 - 2x + 1 + 1)  = - x2 + 2x - 2 b)  Check the polynomial for the roots, Quadratic function or the second-degree polynomial
The polynomial function of the second degree,  f (x) = a2x2 + a1x + a0 is called a quadratic function. y = f (x)  = a2x2 + a1x + a0   or   y - y0 = a2(x - x0)2,
 where are the coordinates of translations of the quadratic
function. By setting   x0 = 0 and  y0 = we obtain  y = a2x2,  the source quadratic function.
The turning point  V(x0, y0) is called the vertex of the parabola.
Note that the coefficients, a2, a1 and a0, of quadratic function, correspond to the coefficients, a, b and c, of
 The real zeros of the quadratic function: The above formula is known quadratic formula that shows the symmetry of the roots relative to the axis of
symmetry of the parabola.
 y = f (x) = a2x2 + a1x + a0  = a2(x - x1)(x - x2) = a2[x2 - (x1 + x2)x + x1x2]
The graph of a quadratic function is curve called a parabola. The parabola is symmetric with respect to a vertical line called the axis of symmetry.
As the axis of symmetry passes through the vertex of the parabola its equation is x = x0.
Quadratic function has the y-intercept at the point ( 0, a0 ).
The proof that quadratic function  f (x) = a2x2 + a1x + a0 is translation of its source or original  f (x) = a2x2
1)  Let calculate the coordinates of translations of quadratic function using the formulas,
 substitute n = 2 in then 2)  To get the source quadratic function we should plug the coordinates of translations (with changed signs)
into the general form of the quadratic, i.e., after expanding and reducing obtained is
y = a2x2   the source quadratic function
3)  Inversely, by plugging the coordinates of translations into the source quadratic function
y - y0 = a2(x - x0)2, and after expanding and reducing we obtain
y = a2x2 + a1x + a0   the quadratic function in the general form.   Intermediate algebra contents 