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FACTORING
Factoring Quadratic Trinomial
if a > 1
Example 1. Factor 2x2 – x – 6.
Steps Solution
1. Multiply the coefficient of x2
and the constant term.
coefficient of x2 = 2
constant term = -6
(2).(-6) = -12
2. Find the factors of -12 that
gives you the sum of the
coefficient of x.
coefficient of x = -1
-12 = (-4).(3)
Steps Solution
3. Copy the expression and
replace the middle term as the
value obtain in step 2.
= 2x2 – x – 6
= 2x2 – 4x + 3x - 6
4. Factor using groupings. (2x2 – 4x) + (3x – 6)
= 2x (x – 2) + 3 (x – 2)
= (2x + 3) (x – 2)
Example 2. Solve for the roots of 10x2 + 3x - 1.
Steps Solution
1. Multiply the coefficient of x2
and the constant term.
coefficient of x2 = 10
constant term = -1
(10) . (-1) = -10
2. Find the factors of -10 that
gives you the sum of the
coefficient of x.
coefficient of x = 3
3 = (5) . (-2)
Steps Solution
3. Copy the expression and
replace the middle term as the
value obtain in step 2.
= 10x2 + 3x – 1
= 10x2 + 5x – 2x – 1
4. Factor using groupings. = (10x2 + 5x) – (2x – 1)
= 5x(2x + 1) – 1(2x + 1)
= (5x – 1)(2x + 1)
Example 3. Find the roots of 3x2 - x – 2.
Steps Solution
1. Multiply the coefficient of x2
and the constant term.
coefficient of x2 = 3
constant term = -2
(3) . (-2) = -6
2. Find the factors of –6 that
gives you the sum of the
coefficient of x.
coefficient of x = -1
(-1) = (-3) . (2)
Steps Solution
3. Copy the expression and
replace the middle term as the
value obtain in step 2.
= 3x2 – x – 2
= 3x2 – 3x + 2x – 2
4. Factor using groupings. = (3x2 – 3x) + (2x – 2)
= 3x(x – 1) + 2(x – 1)
= (3x + 2) (x – 1)
ThankYou!

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Factoring if a is greater than 1 grade8

  • 3. Example 1. Factor 2x2 – x – 6. Steps Solution 1. Multiply the coefficient of x2 and the constant term. coefficient of x2 = 2 constant term = -6 (2).(-6) = -12 2. Find the factors of -12 that gives you the sum of the coefficient of x. coefficient of x = -1 -12 = (-4).(3)
  • 4. Steps Solution 3. Copy the expression and replace the middle term as the value obtain in step 2. = 2x2 – x – 6 = 2x2 – 4x + 3x - 6 4. Factor using groupings. (2x2 – 4x) + (3x – 6) = 2x (x – 2) + 3 (x – 2) = (2x + 3) (x – 2)
  • 5. Example 2. Solve for the roots of 10x2 + 3x - 1. Steps Solution 1. Multiply the coefficient of x2 and the constant term. coefficient of x2 = 10 constant term = -1 (10) . (-1) = -10 2. Find the factors of -10 that gives you the sum of the coefficient of x. coefficient of x = 3 3 = (5) . (-2)
  • 6. Steps Solution 3. Copy the expression and replace the middle term as the value obtain in step 2. = 10x2 + 3x – 1 = 10x2 + 5x – 2x – 1 4. Factor using groupings. = (10x2 + 5x) – (2x – 1) = 5x(2x + 1) – 1(2x + 1) = (5x – 1)(2x + 1)
  • 7. Example 3. Find the roots of 3x2 - x – 2. Steps Solution 1. Multiply the coefficient of x2 and the constant term. coefficient of x2 = 3 constant term = -2 (3) . (-2) = -6 2. Find the factors of –6 that gives you the sum of the coefficient of x. coefficient of x = -1 (-1) = (-3) . (2)
  • 8. Steps Solution 3. Copy the expression and replace the middle term as the value obtain in step 2. = 3x2 – x – 2 = 3x2 – 3x + 2x – 2 4. Factor using groupings. = (3x2 – 3x) + (2x – 2) = 3x(x – 1) + 2(x – 1) = (3x + 2) (x – 1)