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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 solutions of the equation w2 + 8w - 33 = 0 are w = -11 or ___ Answer: Question 2: This question is available to subscribers only!