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Middle/High School Algebra, Geometry, and Statistics (AGS)
4.5 Factoring Quadratics with One Variable

  • A quadratic expression or equation is a polynomial whose highest exponent is 2.
  • The root root is the solution to the quadratic equation.
  • Using factoring we find the solution of the quadratic equation.
  • A quadratic will have a double root if the quadratic is a perfect square trinomial.

Factoring x2 + bx + c
x2 + bx + c = (x + p) (x + q) provided p + q = b and pq =c.

When factoring a trinomial, first consider the signs of p and q.
x2+bx+c(x+2)(x+3) (x+2)(x+(-3)) (x+(-2))(x+3) (x+(-2))(x+(-3))

(x + p)(x + q)
signs of b and c
x2+5x+6 b is positive; c is positive
x2-x-6 b is negative; c is negative
x2+5x-6 b is positive; c is negative
x2-5x+6 b is negative; c is positive

Observing the signs of b and c in the table we see that:
* b and c are positive when both p and q are positive.
* b is negative and c is positive when both p and q are negative.
* c is negative when p and q have different signs.

Examples:

  1. Factor x2 + 11x + 18
    Find two positive factors of 18 whose sum is 11. Make a table
    Factors of 18 Sum of factors
    18, 1 18+1=19 Not correct
    9,2 9 + 2 = 11 Correct sum
    6, 3 6 + 3 = 9 Not correct

    The factors 9 and 2 have a sum of 11, so they are correct values of p and q.
    x2 + 11x + 18 = (x + 9)(x + 2)

  2. Show that the factors of n2 - 6n + 8 is (n-4)(n-2)
    Factors of 8 Sum of factors
    -8, -1 -8+(-1)=-9 Not correct
    -4,-2 -4 + (-2) = -6 Correct sum

    n2 - 6n + 8 = (n-4)(n-2)

  3. Show that the factors of y2 + 2y - 15 = (y + 5)(y - 3)
    Factors of -15 Sum of factors
    -15, 1 -15+1=-14 Not correct
    15, -1 15-1=14 Not correct
    -5, 3 -5+3=-2 Not correct
    5, -3 5-3=2 Correct sum

    y2 + 2y - 15 = (y + 5)(y - 3)

  4. Solve the equation x2 + 3x = 18 x2 + 3x = 18 ----- original equation
    x2 + 3x - 18 = 0 ----- subtract 18 from each side
    (x + 6)(x - 3) = 0 ----- factor left side
    (x + 6) = 0 or x - 3 = 0
    Therefore x = -6 or x = 3 ------ solve for x
    The solutions of the equation are -6 and 3

  5. Solve 6x2 + 42x = 0
    6x2 + 42x = 0 ---- original equation
    6x(x + 7) = 0 --- factor left side
    6x = 0 or x + 7 = 0 ---- solve for x
    x = 0 or x = -7
    The solutions of the equation are 0 and -7

Directions: Solve the following problems. Also write at least ten examples of your own.
Q 1: The root of the quadratic equation: 3x2 + 14x - 5 = 0 are x = -5 and x = __
Answer:

Q 2: The quadratic x2 - 10x + 25 can be factored as (x-5)(x-_)
Answer:

Q 3: The root of the quadratic equation: 2x2 - 7x + 3 = 0 are x = 3 and x = __
Answer:

Q 4: The root of the quadratic equation: x2 + 3x - 4 = 0 are x = __ and x = 1
Answer:

Q 5: The root of the quadratic equation: -16a2 + 46a + 6 = 0 are a = -1/8 and t = __
Answer:

Q 6: Solve the quadratic equation: x2 = x + 20
x = __ or 5
Answer:

Q 7: The root of the quadratic equation: 3- (5/2)x - 3x2 = 0 are x = __ and x = -3/2
Answer:

Q 8: The solutions of the equation w2 + 8w - 33 = 0 are w = -11 or ___
Answer:

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Question 10: This question is available to subscribers only!


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