ALGEBRA - solved problems
Polynomial and/or polynomial functions and equations
 126
Solution:   a( - 5x3 + 2x2 - x + 4) + ( - 4x2 + 3x - 7) = - 5x3 + ( 2 - 4) · x2  + ( -1 + 3) · x + 4 - 7
= - 5x3 - 2x2  + 2x - 3.
b( x4 - 3x3 + 5x - 1) - ( - 2x4 + x3 - 3x2 + 4) = x4 - 3x3 + 5x - 1 + 2x4 - x3 + 3x2 - 4
= ( 1 + 2) · x4  + ( -3 - 1) · x3 + 3x2 + 5x - 5 = 3x - 4x3 + 3x2 + 5x - 5.
Multiplication of polynomials
 127
Multiply given polynomials.
Solution:    ( - 2x3 + 5x2 - x + 1) · ( 3x - 2) =
= 3x · ( - 2x3) + 3x · 5x2 + 3x ·  (-x) + 3x · 1 + (- 2) · ( - 2x3) + (- 2) · 5x2 + (- 2) · ( -x) + (- 2) · 1
= - 6x4 + 15x3 - 3x2 + 3x + 4x3 - 10x2 + 2x - 2 = - 6x4 + 19x3 - 13x2 + 5x - 2.
Division of polynomials
 128
Divide given polynomials.
 Solution:
Note, since each second line should be subtracted, the sign of each term is reversed.
 129
Divide the polynomials (  3x4 + x2 + 5) ¸ ( x2 - x - 1).
 Solution:
 or
 like 17 ¸  5 = 3 + 2/5 -15 2
 130
Solve the quadratic equation  ax2 + bx + c = 0 by completing the square, derive the quadratic
formula.
 Solution:
the roots can also be written,
 131
A train made up for delay of 12 minutes after 60 km of way by running 10 km/h faster then regular
speed.  What is the regular speed of the train?
 Solution:
 132
In a theater each row has the same number of seats. Number of rows equals number of seats.
By doubling number of rows and decreasing number of seats 10 per row, total number of seats in the theater increases by 300. How many rows are in the theater?
Solution:   Taking x as the number of rows (or seats per row), then
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.
 133
Find the roots of each quadratic equation by factoring.
 Solution: 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,     and      x + 2 = 0    =>      x2 = -2.
 b)  2x2 - 7x + 3 = 0 since 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,     and      x - 3 = 0    =>    x2 = 3.
 c)  3x2 - x - 2 = 0 since 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.
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