Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Section 5.4 Factoring FACTORING Greatest Common Factor, Factor by Grouping, Factoring Trinomials, Difference of Squares, Perfect Square Trinomial, Sum & Difference of Cubes 1 Factoring—define factored form Factor means to write a quantity as a multiplication problem a product of the factors. z Factored forms of 18 are: z 1i18, 2i9, 3i6 2 Factoring: The Greatest Common Factor z 1. 2. 3. z To find the greatest common factor of a list of variable terms: Choose the variables common to each term. Choose the smallest exponent of each common variable. Multiply the variables. Find the GCF of: x 4 y 2 , x 7 y , x3 y 5 , x 2 y 6 = 3 x2 y Factoring: The Greatest Common Factor z 1. 2. 3. 4 To factor out the greatest common factor of a polynomial: Choose the greatest common factor for the coefficients. Choose the greatest common factor for the variable parts. Multiply the factors. Factoring: The Greatest Common Factor Factor each of the following by factoring out the greatest common factor: z 5 5x + 5 = 4ab + 10a2 = 8p4q3 + 6p3q2 = 2y + 4y2 + 16y3 = 3x(y + 2) -1(y + 2) = Factoring: The Greatest Common Factor z The answers are : 5 x + 5 = 5 ( x + 1) 4ab + 10a 2 = 2a ( 2b + 5a ) 8 p 4 q 3 + 6 p 3 q 2 = ( 2 p 3 q 2 ) ( 4 pq + 3) 2 y + 4 y + 16 y = 2 y (1 + 2 y + 8 y 2 3 2 ) 3 x( y + 2 ) − 1( y + 2 ) = ( y + 2 )( 3 x − 1) 6 Factoring: by Grouping z 1. 2. 3. 7 Often used when factoring four terms. Organize the terms in two groups of two terms. Factor out the greatest common factor from each group of two terms. Factor out the common binomial factor from the two groups. Rearranging the terms may be necessary. Factoring: by Grouping z Factor by grouping: 1. 2 groups of 2 terms Factor out the GCF from each group of 2 terms 2n ( m − 4 ) + 3 ( m − 4 ) Factor out the common binomial factor ( m − 4 )( 2n + 3) 2. 3. 8 2mn − 8n + 3m − 12 Factoring: by Grouping Factor by grouping 6 y 2 − 20w2 + 15 yw − 8 yw ( ) 2 3 y 2 − 10w2 + ???????? rearrange the terms and try again 6 y 2 − 8 yw + 15 yw − 20w2 2 y ( 3 y − 4 w ) + 5w ( 3 y − 4 w ) (3 y − 4w)( 2 y + 5w) 9 Factoring Trinomials—with a coefficient of 1 for the squared term z 1. 2. 3. Factor: x + 12 y + 20 List the factors of 20: Select the pairs from which 12 may be obtained Write the two ( x + 10 )( x + 2 ) binomial factors: 2 x 2 + 2 x + 10 x + 20 4. 10 Check using FOIL: x + 12 x + 20 2 20 1i 20 2i10 4 i5 Factoring Trinomials ÍTIPÍ If the last term of the trinomial is positive and the middle sign is positive, both binomials will have the same “middle” sign as the second term. x + 12 x + 20 2 ( x + 10 )( x + 2 ) 11 Factoring Trinomials ÍTIPÍ If the last term of the trinomial is positive and the middle sign is negative, both binomials will have the same “middle” sign as the second term. x − 12 x + 20 2 ( x − 10 )( x − 2 ) 12 Factoring Trinomials—with a coefficient of 1 for the squared term z 1. 2. 3. 4. 13 Factor a − 9a − 22 22 List the factors of 22 1 i 2 2 Select the pair from 2i11 which –9 may be obtained Write the two ( x − 11)( x + 2 ) binomial factors: 2 Check using FOIL: x 2 + 2 x − 11x − 22 x 2 − 9 x − 22 Factoring Trinomials ÍTIPÍ If the last term of the trinomial is negative, both binomials will have one plus and one minus “middle” sign. x − 9 x − 22 2 ( x − 11)( x + 2 ) 14 Factoring Trinomials—primes A PRIME POLYNOMIAL cannot be factored using only integer factors. 2 z Factor : x − 2 x + 5 z The factors of 5: 1 and 5. z Since –2 cannot be obtained from 1 and 5, the polynomial is prime. z 15 Factoring Trinomials—2 variables y 2 − 6 yz + 8 z 2 Factor: z The factors of 8 are: 1,8 & 2,4, & -1,-8 & -2, -4 z Choose the pairs from which –6 can be obtained: 2 & 4 z Use y in the first y − 4z y − 2z position and z in the second position 2 2 y − 2 yz − 4 yz + 8 z z Write the two binomial factors and 2 2 y − 6 yz + 8 z check your answer z ( 16 )( ) Factoring Trinomials—with a GCF z z If there is a greatest common factor? If yes, factor it out first. Factor: 3 z 4 − 15 z 3 + 18 z 2 3z 2 (z 3z 2 ( z − )( z − ) 2 − 5 z + 6 ) factors of 6: 1i6 & 2i3 Choose 2i3 3 z 2 ( z − 2 )( z − 3) 17 Factoring Trinomials—always check your factored form z Always check your answer with multiplication of the factors. z The check: 3z 2 3z 2 3z 2 ( z − 2 )( z − 3) (z (z 2 − 3z − 2 z + 6 ) 2 − 5z + 6) 3 z − 15 z + 18 z 4 18 3 3 Factoring Trinomials— when the coefficient is not 1 on the squared term z Factor: 3x 2 − 4 x + 1 1. Multiply 3i1 = 3 2. List the factors of 3: 1i3 3. Rewrite the middle term of the trinomial using the factors of 3. 3x 2 − x − 3x + 1 4. Factor by grouping. x ( 3 x − 1) − 1( 3 x − 1) = 19 ( x − 1)( 3x − 1) Factoring Trinomials---use grouping z Factor: 6x2 + x −1 1. Multiply 6i1 = 6 2. List the factors of 6: 1i6 & 2i3 3. Rewrite the middle term of the trinomial using the factors of 6 which add to be 1 ie. 2i3. 6x2 + 3x − 2x −1 4. Factor by grouping. 3x ( 2x +1) −1( 2x +1) = 20 ( 2x +1)( 3x −1) Factoring Trinomials---use grouping z Factor: 12x 2 − 5 x − 2 1. Multiply 12i2 = 24 2. List the factors of 24: 1i24; 2i12; 3i8; & 4i6 3. Rewrite the middle term of the trinomial using the factors of 24 which add to be 5 ie. 3i8. 12 x 2 − 8 x + 3 x − 2 4. Factor by grouping. 4 x ( 3 x − 2 ) + 1( 3 x − 2 ) = 21 ( 3x − 2 )( 4 x + 1) Factoring Trinomials---use FOIL and Trial and Error z Factor: 6 x 2 + 19 x + 10 There will be two binomial factors. ( Both middle signs will be positive. The factors of 6 are 1 • 6 and 2 • 3. The factors of 10 are 1 • 10 and 2 • 5. (Continued on next screen.) 22 )( ( + ) )( + ) Factoring Trinomials---use FOIL and Trial and Error z Think FOIL and focus on the outers and inners while using trial and error to position the factors correctly. ( 2x + 5)( 3x + 2 ) inners: 15x outers: 4x Always check your answer. 23 2 2 x + 5 3 x + 2 = 6 x + 19 x + 10 ( )( ) Factoring Trinomials---use FOIL and Trial and Error z Factor: 10x 2 − 23x + 12 There will be two binomial factors. ( Both middle signs will be negative. )( ( − The factors of 10 are 1 • 10 and 2 • 5. The factors of 12 are 1 • 12, 2 • 6, and 3 • 4. (Continued on next screen.) 24 ) )( − ) Factoring Trinomials---use FOIL and Trial and Error z Think FOIL and focus on the outers and inners while using trial and error to position the ( 2x − 3)( 5x − 4 ) factors correctly. inners: − 15x outers: − 8 x Always check your answer. 25 ( 2x − 3)( 5 x − 4 ) = 10 x 2 − 23 x + 12 Factoring Trinomials---use FOIL and Trial and Error 2 Factor: 5 x + 13 x − 6 z There will be two binomial factors. One middle sign will be negative. One middle sign will be positive. The factors of 5 are 1 • 5. The factors of 6 are 1 • 6, and 2 • 3. (Continued on next screen.) 26 ( ( − )( )( ) + ) Factoring Trinomials---use FOIL and Trial and Error z Think FOIL and focus on the outers and inners while using trial and error to position the factors correctly. ( 5 x − 2 )( x + 3) inners: − 2 x outers: 15x Always check your answer. 27 2 5 x − 2 x + 3 = 5 x + 13 x − 6 ( )( ) Factoring Trinomials---with a negative GCF z z z Is the squared term negative? If yes, factor our a negative GCF. Factor: − 4a 2 + 2a + 30 − 2a ( 2a 2 − a − 15 ) − 2a ( 2a + 5 )( a − 3) The sum of inners: 5a and outers: − 6a is − a 28 Special Factoring—difference of 2 squares 9 The following must be true: 1. There must be only two terms in the polynomial. Both terms must be perfect squares. There must be a “minus” sign between the two terms. 2. 3. 29 Special Factoring—difference of 2 squares z The following pattern holds true for the difference of 2 squares: z 30 x − y = ( x + y )( x − y ) 2 2 Special Factoring—difference of 2 squares z z The pattern: x 2 − y 2 = ( x + y )( x − y ) Factor: x 2 − 25 x 2 is a perfect square since x2 = x 25 is a perfect square since 25 = 5 Use the pattern letting y = 5. x − ( 5 ) = ( x + 5 )( x − 5 ) 2 31 2 Special Factoring—difference of 2 squares z z The pattern: x 2 − y 2 = ( x + y )( x − y ) Factor: a − b 2 2 a 2 is a perfect square since a 2 = a b 2 is a perfect square since b 2 = b Use the pattern letting x = a & y = b. a − b = ( a + b )( a − b ) 2 32 2 Special Factoring—difference of 2 squares z z The pattern: x 2 − y 2 = ( x + y )( x − y ) Factor: 9 x − 4 y 2 2 9 x 2 is a perfect square since 9 x 2 = 3 x 4 y 2 is a perfect square since 4 y 2 = 2 y Use the pattern letting x = 3 x & y = 2 y. ( 3x ) − ( 2 y ) 2 33 2 = ( 3x + 2 y )( 3 x − 2 y ) Special Factoring—difference of 2 squares z z The pattern: x 2 − y 2 = ( x + y )( x − y ) Factor: 9 x + 4 y 2 2 Although both terms are perfect squares, the pattern does not apply because the problem is a sum and not a difference. 34 Special Factoring—perfect square trinomial z A perfect square trinomial is a trinomial that is the square of a binomial. z 4 x 2 + 4 xy + y 2 is a perfect square trinomial because: ( 2 x + y ) = ( 2 x + y )( 2 x + y ) = 2 2 2 2 ( 2 x ) + 2 xy + 2 xy + y = 4 x + 4 xy + y 2 35 Special Factoring—perfect square trinomial z 4 x2 z z z ¾ 36 + 4 xy + y 2 is a perfect square trinomial because: The first and third terms are perfect squares. AND the middle term is twice the product of the square roots of the first and third terms TEST THE MIDDLE TERM: 4x = 2x 2 y = y 2 ( 2 x )( y ) = 4 xy 2 Special Factoring—perfect square trinomial z The patterns for a perfect square trinomial are: x + 2 xy + y = ( x + y )( x + y ) = ( x + y ) 2 2 x − 2 xy + y = ( x − y )( x − y ) = ( x − y ) 2 37 2 2 2 Special Factoring—perfect square trinomial z Factor the following using the perfect square trinomial pattern: 16a 2 + 56a + 49 = ( 4a + 7 ) 2 The first and third terms are perfect squares. 16a 2 = 4a 49 = 7 The middle term is twice the product of the square root of the first and third terms. 2 ( 4a )( 7 ) = 56a 38 Special Factoring—perfect square trinomial z Factor the following using the perfect square trinomial pattern: x − 22 xz + 121z = ( x − 11z ) 2 2 2 The first and third terms are perfect squares. x2 = x 121z 2 = 11z The middle term is twice the product of the square root of the first and third terms. 2 ( x )(11z ) = 22 xz 39 Special Factoring—difference of 3 3 2 2 two cubes x − y = ( x − y ) ( x + xy + y ) z Factor using the pattern. 27 x3 − 8 y 3 Both terms must be perfect cubes. 3 27x3 = 3 x; 3 8y 3 = 2 y ( 3x ) − ( 2 y ) Let x = 3x, Let y = 2 y 2 2 ( 3x − 2 y ) ( ( 3x ) + ( 3x )( 2 y ) + ( 2 y ) ) 3 3 2 2 3 x − 2 y 9 x + 6 xy + 4 y ( )( ) 40 Special Factoring—sum of two cubes x3 + y 3 = ( x + y ) ( x 2 − xy + y 2 ) z Factor using the pattern. 1000 x3 + 125 y 3 Both terms must be perfect cubes. 3 1000 x 3 = 10 x; 3 125 y 3 = 5 y (10 x ) + ( 5 y ) Let x = 10 x, Let y = 5 y 2 2 10 5 10 10 5 5 x y x x y y + − + ( ) ( ( ) ( )( ) ( ) ) 3 3 2 2 10 5 100 50 25 x + y x − xy + y ( )( ) 41 Solving quadratic equation with factoring zA quadratic equation has a “squared” term. ax + bx + c = 0 2 42 ZERO FACTOR PROPERTY To Factor a Quadratic, Apply the Zero-Factor Property. ¾ If a and b are real numbers and if ab = 0, then a = 0 or b = 0. 43 Solving quadratic equations with factoring—Zero-Factor Property z Solve the equation: (x + 2)(x - 8) = 0. z Apply the zero-factor property: (x + 2) = 0 x = -2 44 or or (x – 8) = 0 x=8 Solving quadratic equations with factoring—Zero-Factor Property There are two answers for x: -2 and 8. z Check by substituting the values calculated for x into the original equation. (x + 2)(x - 8) = 0. (-2 + 2)(-2 – 8) = 0 (8 + 2)(8 – 8) = 0 0=0 0=0 z 45 Solving quadratic equations with factoring—Standard Form z 1. To solve a quadratic equation, Write the equation in standard form. ¾ (Solve the equation for 0.) x + 6 x = −8 ⇒ x + 6 x + 8 = 0 2 46 2 Solving quadratic equations with factoring z 2. To solve a quadratic equation, Factor the quadratic expression. x + 6x + 8 = 0 2 ( x + 2 )( x + 4 ) = 0 47 Solving quadratic equations with factoring z 3. To solve a quadratic equation, Apply the Zero-Factor Property x + 6x + 8 = 0 2 ( x + 2 )( x + 4 ) = 0 x+2=0 x+4=0 x = −2 x = −4 48 Solving quadratic equations with factoring z 4. To solve a quadratic equation, Check your answers x2 + 6 x + 8 = 0 x = −2, or x = −4 ( −2) ( −4) 2 + 6 ( −2) + 8 = 0 4 − 12 + 8 = 0 −8 + 8 = 0 0=0 49 2 + 6 ( −4 ) + 8 = 0 16 − 24 + 8 = 0 −8+8 = 0 0=0