ALGEBRA - solved problems
Factoring algebraic expressions
Factoring algebraic expression by finding (determining) a common factor
 84 Factorize given expressions.
 Solutions: a)  3x - 6y = 3 · (x - 2y),     b)  xy - y2  = y · (x - y),     c)  a - a2  = a · (1 - a), d)  x3 -3x2  + x = x · (x2 - 3x +1),    e)  x(a + b) - (a + b) = (a + b) · (x - 1), f)   a(x - 3y) - x + 3y  = a(x - 3y) - (x - 3y) = (x - 3y) · (a - 1).
Grouping like terms, grouping and factorizing four terms
 85 Factorize given expressions.
 Solutions: a)  ax - bx - a + b = x(a - b) - (a - b) = (a - b) · (x - 1), b)  a - 1 - ab + b = (a - 1) - b · (a - 1) = (a - 1) · (1 - b), c)  x2 + ax - bx - ab = x(x + a) - b · (x + a) = (x + a) · (x - b), d)  5ab2 - 3a3 - 10b3 + 6a2b = 5b2(a - 2b) -3a2(a - 2b) = (a - 2b)(5b2 - 3a2).
The square of a binomial - perfect squares trinomials
 86 Factorize given expressions.
 Solutions: a)  1 - 4x + 4x2 = 12  - 2 · 2x + (2x)2 = (1 - 2x)2  = (1 - 2x) · (1 - 2x), b)  a5 + 6a4b + 9a3b2 = a3 · (a2  + 6ab  + 9b2 ) = a3(a + 3b)2 = a3(a + 3b)(a + 3b).
Difference of two squares
 87 Factorize given expressions.
 Solutions: a)  16x2 - 1 = (4x)2 - 12 = (4x -1) · (4x +1), b)  5y3 - 20x2y = 5y · (y2 - 4x2) = 5y [y2  - (2x)2] = 5y(y - 2x)(y + 2x), c)   9x2  - (x + 2)2 = [3x - (x + 2)] · [3x + (x + 2)] = (2x -2) · (4x + 2) = 4(x -1) · (2x +1).
Factoring quadratic trinomials
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
 88 Factorize quadratic trinomials.
 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).
 89 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].
 90 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.
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