ALGEBRA - solved problems
Polynomial and/or polynomial functions and equations
Factoring polynomials and solving polynomial equations by factoring
A polynomial and/or polynomial function with real coefficients can be expressed as a product of its leading coefficient an and n linear factors of the form x - xi, where xi denotes its real roots and/or complex roots,
f (x) = anxn + an-1xn-1 + . . . + a1x + a0 = an(x - x1)(x - x2) . . . (x - xn).
Thus, a quadratic or a second degree polynomial
a2x2 + a1x + a0 = a2(x - x1)(x - x2) = a2[x2 - (x1 + x2)x + x1x2],
and similarly a cubic or a third degree polynomial
a3x3 + a2x2 + a1x + a0 = a3(x - x1)(x - x2)(x - x3) =
= a3[x3 - (x1 + x2 + x3)x2 + (x1x2 + x1x3 + x2x3)x - x1x2x3]
and so on.
 134
Factorize  (2/3)x2 - (2/3)x - using the above theorem.
Solution:    (2/3)x2 - (2/3)x - 4 = (2/3)(x2 - x - 6) = (2/3)[x2 - (3 + (- 2))x + 3(- 2)] =
= (2/3)(x2 - 3x + 2x - 6) = (2/3)[x(x - 3) + 2(x - 3)] = (2/3)(x - 3)(x + 2).
 135
Given are leading coefficient a2 = -1and the pair of conjugate complex roots, x1 = 1 + and
x2 = 1 - i, of a second degree polynomial, find the polynomial using the above theorem.
Solution:    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
 136
Check the roots of the quadratic found in the previous problem using the quadratic formula.
 Solution:
 137
The real root of the polynomial  - x3 - x2 + 4x - 6 is x1 = - 3, factorize the polynomial.
Solution:    We divide given polynomial by one of its known factors, since
a3x3 + a2x2 + a1x + a0 = a3(x - x1) (x - x2) (x - x3)
then we calculate another two roots of given cubic by solving obtained quadratic trinomial,
Finally we use the theorem to factorize given polynomial (see the previous example),
a3(x - x1)(x - x2)(x - x3) = -1(x + 3)[x - (1 + i)][x - (1 - i)] = -(x + 3)(x2 - 2x + 2).
Notice that given cubic has one real root and the pair of the conjugate complex roots.
Odd degree polynomials must have at least one real root.
 138
Solve the cubic equation  x3  +  2x2 - x - 2 = 0 by factoring.
Solution:            x2(x +  2)  - (x +  2) = 0
(x +  2)·(x2 -  1) = 0
x +  2 = 0      =>    x1 = -2
x2 -  1 = 0     =>       x +  1 = 0     =>     x2 = -1
x - 1 = 0    =>     x3 = 1
Therefore, the roots are,    x1 = -2,    x2 = -1   and    x3 = 1.
Polynomial functions
Transformations of the polynomial function applied to the quadratic and cubic functions
y = f (x = a2x2 + a1x + a0    or    y - y0 = a2(x - x0)2,
 where coordinates of translations of the quadratic function.
By setting   x0 = y0 = 0,   we get  y = a2x2,  the source quadratic function.  The turning point  V(x0, y0).
 The real zeros of the quadratic function
 139
Find zeros and vertex of the quadratic function  y = - x2 + 2x + 3  and sketch its graph.
Solution:   A quadratic function can be rewritten into translatable form  y - y0 = a2(x - x0)2  by completing
the square,
 y = - x2 + 2x + 3 Since a2 · y0 < 0 given quadratic function must have two different real zeros. y = - (x2 - 2x) + 3 To find zeros of a function, we set y equal to zero and solve for x. Thus, y = - [(x - 1)2 - 1] + 3 - 4 = - (x - 1)2 y -  4 = - (x - 1)2 (x - 1)2 = 4 y - y0 = a2(x - x0)2 x - 1 = + sqrt(4) V(x0, y0)  =>   V(1, 4) x1,2 = 1 + 2,   =>   x1 = - 1 and  x2 = 3.