ALGEBRA - solved problems
Algebraic expressions
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
 Solutions: a)  x2 - 3x -10 = x2 + (-5 + 2)·x + (-5)·(+2) = x2 - 5x + 2x -10 = = x · (x - 5) + 2 · (x - 5) = (x - 5) · (x + 2), b)  2x2 - 7x + 3 = 2 · (x2 - 7/2x + 3/2) =  2(x2 - 1/2x - 3x + 3/2) = = 2[x·(x - 1/2) - 3· (x - 1/2)] = 2· (x - 1/2)·(x - 3) = (2x - 1)· (x - 3), c)  3x2 - x - 2 = 3(x2 - 1/3x - 2/3) =  3(x2 + 2/3x - x - 2/3) = = 3[x·(x + 2/3) - (x + 2/3)] = 3·(x + 2/3)·(x - 1) = (3x + 2)·(x - 1).
 12 Given are leading coefficient a2 = -1 and the complex roots,  x1 = 1 + i  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.
Factoring 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].
 13 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, 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.
Sum and difference of cubes
 14 Factorize given sum and difference of cubes.
 Solutions: a)  x3 + 8 = x3 + 23 = (x + 2) · (x2 - 2x + 22), since  (x + 2)·(x2 - 2x + 4) = x3 - 2x2 + 4x + 2x2 - 4x + 8 = x3 + 8, b) 8a3 -125 = (2a)3 - 53 = (2a - 5)· [(2a)2 + (2a)·5 + 52] = (2a - 5)(4a2 + 10a + 25), since  (2a - 5)(4a2 + 10a + 25) = 8a3 + 20a2 + 50a - 20a2 - 50a -125 = 8a3 -125.
Using a variety of methods including combinations of the above to factorize expressions
 15 Factorize given expressions.
 Solutions: a)  x2 - 2xy + y2 + 2y - 2x = (x - y)2 - 2(x - y) = (x - y)(x - y - 2), b)  x2 - y2 + xz - yz = (x - y)(x + y) + z(x - y) = (x - y)(x + y + z), c)  4x2  - 4xy  + y2  - z2 = (2x - y)2   - z2 = (2x - y - z)(2x - y + z), d)  a3  - 7a + 6 = a3  - a - 6a + 6 = a(a2 -1) - 6(a -1) = (a -1)·[a(a + 1) - 6] = (a -1)(a2 + a - 6) = (a -1)(a2 + 3a - 2a - 6) = (a -1)[a(a + 3) - 2(a + 3)] = (a -1)(a + 3)(a - 2).
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