Factor TriNomials - Intermediate Algebra - Lecture Slides, Slides of Algebra

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§5.4 Factor

TriNomials

Review §

 Any QUESTIONS About

• §5.3 → Factoring by GCF and/or Grouping

 Any QUESTIONS About HomeWork

• §5.3 → HW-

5.3 MTH 55

Factor (+1)· x^2 + bx + c

• To factor x^2 + 7 x + 10, think of FOIL: The first term, x^2 , is the product of the First terms of two binomial factors, so the first term in each binomial must be x.
• The challenge is to find two numbers p and q such that p • q = c , and p + q = b

x^2 + 7 x + 10 = ( x + p )( x + q )

= x^2 + qx + px + pq

Factor (+1)· x^2 + bx + c

• Need to find two numbers p & q such that

x^2 + 7 x + 10 = ( x + p )( x + q ) = x^2 + qx + px + pq

 Thus the numbers p and q must be selected so that their

• PRODUCT is 10
• SUM is 7  The Factor Pairs for 10 [and their sums]
• 1 ·10 [11]; (−1)·(−10) [−11]; 2·5 [7]; (−2)·(−5) [−7];

Example  FOIL Factoring

• Multiplying binomials uses the FOIL method, Factoring uses the FOIL method backwards Product of x and x is x^2.

Product of 5 and –7 is –35. Sum of the product of outer and inner terms

O I

F

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Factor x^2 + b x + c for Positive c

• When the constant term of a trinomial is

positive , look for two numbers with the

same sign. The sign is that of the middle

term:

x^2 – 7 x + 10 = ( x – 2)( x – 5);

x^2 + 7 x + 10 = ( x + 2)( x + 5);

Example  Factor x^2 + 7 x + 12

• Since 3 ⋅ 4 = 12 and 3 + 4 = 7, the factorization of x^2 + 7 x + 12 is ( x + 3)( x + 4).
• To check we simply multiply the two binomials.
• CHECK by FOIL:

( x + 3)( x + 4) = x^2 + 4 x + 3 x + 12 = x^2 + 7 x + 12 

Example  Factor y^2 – 8 y + 15

• SOLUTION: Since the constant term is positive and the coefficient of the middle term is negative, we look for the factorization of 15 in which both factors are negative. Their SUM must be Pairs of −8. Factors of 15

Sums of Factors –1, –15 – –3, –5 –

Sum of − 8

y^2 − 8 y + 15 = ( y − 3)( y – 5)

Example  Factor x^2 – 5 x – 24

• SOLUTION: The constant term must be expressed as the product of negative & positive numbers.
• Since the sum of the two numbers must be negative, the negative number must have the greater absolute value.

Pairs of Factors of 24

Sums of Factors 1, − 24 − 23 2, − 12 − 10 3, − 8 − 5 4, − 6 − 2 6, − 4 2 8, − 3 5

x^2 − 5 x − 24 = ( x + 3)( x – 8)

Example  Factor t^2 – 32 + 4 t

• SOLUTION:
• Rewrite the trinomial t^2
  • 4 t − 32.
• We need one positive and one negative factor. The sum must be 4, so the positive factor must have the larger absolute value

t^2 + 4 t − 32 = ( t + 8)( t − 4)

Pairs of Factors of 32

Sums of Factors −1, 32 31 −2, 16 14 −4, 8 4

Prime Polynomials

• A polynomial that canNOT be factored is considered prime. - Example: x^2 − x + 7
• Often factoring requires two or more steps. Remember, when told to factor, we should factor completely. This means the final factorization should contain only prime polynomials.

Example  Factor 2 x^3 − 24 x^2 +72 x

• SOLUTION Always look first for a common factor. In this case factor out 2 x : 2 x ( x^2 − 12 x + 36)
• Since the constant term is positive and the coefficient of the middle term is negative, we look for the factorization of 36 in which both factors are negative. - Their SUM must be −12.

To Factor (+1)· x^2 + b x + c

1. Distribute out Common Factors
2. Find a pair of factors that have c as their product and b as their sum. a) If c is positive, its factors will have the same sign as b. b) If c is negative, one factor will be positive and the other will be negative. Select the factors such that the factor with the larger absolute value has the same sign as b.
3. CHECK by MULTIPLYING

Factoring When: LeadCoeff ≠ 1

• Factoring Trinomials of the

Type a x^2 + bx + c

• Factoring with FOIL
• The Grouping Method