VIEWS: 284 PAGES: 25 POSTED ON: 4/15/2011 Public Domain
5.2: Solving Quadratic Equations by Factoring (p. 256) Tucker Method of Factoring Factoring Simple Trinomials Factoring simple polynomials in the form of x2 + bx + c. x2 + 5x + 6 Factors of: +6 An example would be: ◦ x2 + 5x + 6 x x 3 ● 2 -3 ● -2 ◦ First factor the last 6 ● 1 number , +6 (include -6 ● -1 the sign). ◦ Next, write the first term (x2) without the exponent on the top of the bar. Factoring Simple Trinomials Factoring x2 + 5x + 6 x2 + 5x + 6 Factors of: ◦ First factor the last number , +6 +6 (include the sign). x x ◦ Find the factors of that 3 + 2 -3 + -2 = +5 = -5 number (+6), that the +3 +2 6 + 1 -6 + -1 = +7 = -7 sum (+) results in the middle number. (+5). ◦ Place those factors (+3 and +2) in the box under the bar. Factoring Simple Trinomials Factoring x2 + 5x + 6 x2 + 5x + 6 ◦ Now take the left side Factors of: +6 (x + 3) and place that binomial in a parenthesis. x x 3 + 2 = +5 6 + 1 = +7 ◦ Then take the right side +3 +2 (x + 2) and place that binomial in a parenthesis. (x + 3) (x + 2) = x2 + 5x + 6 ◦ You have now factored the polynomial. Factoring Simple Trinomials Now, Factor: x2 - 5x - 6 x2 - 5x - 6 ◦ First factor the last Factors of: number , -6 (include -6 the sign). x x -3 + 2 = -1 ◦ Find the factors of 3 + -2 = +1 that number (-6), -6 +1 6 + -1 = +5 that the difference (-) -6 + 1 = -5 results in the middle number. (-5). ◦ Place those factors (- 6 and +1) in the box under the bar. Factoring Simple Trinomials Factoring x2 - 5x - 6 x2 - 5x - 6 ◦ Now take the left Factors of: -6 side (x - 6) and place that binomial in a x x -3 + 2 = -1 parenthesis. -6 +1 3 6 + + -2 -1 = +1 = +5 ◦ Then take the right -6 + 1 = -5 side (x + 1) and place that binomial in a parenthesis. (x - 6) (x + 1) = x2 - 5x - 6 ◦ You have now factored the polynomial. Factoring Simple Trinomials Factor: x2 - 7x + 12 x2 - 7x +12 ◦ First factor the last Factors of: number , +12 +12 (include the sign). x x 2 + 6 = +8 ◦ Find the factors of -2 + -6 = -8 that number (+12), -3 -4 3 + 4 = +7 that the sum (+) -3 + -4 = -7 results in the middle 12 + 1 = +13 number. (-7). -6 + 1 = -5 ◦ Place those factors (- 3 and -4) in the box under the bar. Factoring Simple Trinomials Factor: x2 - 7x + 12 x2 - 7x + 12 ◦ Now take the left Factors of: +12 side (x - 3) and place that binomial x x 2 + 6 = +8 in a parenthesis. -2 + -6 = -8 3 + 4 = +7 ◦ Then take the right -3 -4 -3 + -4 = -7 side (x - 4) and place 12 + 1 = +13 -6 + 1 = -5 that binomial in a parenthesis. (x - 3) (x - 4) = x2 - 7x + 12 ◦ You have now factored the polynomial. Factoring Trinomials ax2 + bx + c Factoring trinomials in the 5x2 - x -18 form of: ax2 + bx + c. Factors of: 5 ● -18 = -90 An example would be: 5x 5x ◦ 5x2 - x –18 1 ● 2 ● -90 -45 ◦ First, place the first term on 3 ● -20 5 ● -18 both sides of the bars. 6 ● -15 9 ● (without the exponent) 10 ● -10 -9 ◦ Second, multiply the first Etc…… (5) and the last number (-18). ◦ Find the factors of -90. Factoring Trinomials ax2 + bx + c Factoring trinomials in the 5x2 - x -18 form of: ax2 + bx + c. Factors of: -90 Factor: 5x2 - x –18 5x 5x 1 + -90 = -89 ◦ After finding the factors, 2 + -45 = - 44 choose the set whose 9 -10 3 + -20 = -17 5 + -18 = - 13 difference equals the 6 + -15 = -9 middle number, -1. 9 + -10 = -1 (Remember that –x is the 10 + -9 = +1 same as –1x). Etc…… ◦ Place that set with signs under the bar. Factoring Trinomials ax2 + bx + c Factoring trinomials in the 5x2 - x -18 form of: ax2 + bx + c. Factors of: -90 To finish factoring you 5x 5x must reduce each side of 1 + -90 = -89 2 + -45 = - 44 the bar if possible. 9 -10 3 + -20 = -17 5 + -18 = - 13 The 5x over –10 will 1x 6 + -15 = -9 2 9 + -10 = -1 reduce to 1x (or x) over - 10 + -9 = +1 2. Etc…… (5x + 9) (x - 2) = 5x2 - x –18 Now, move each side into the factored answer. Factoring Trinomials ax2 + bx + c Another example would be: 9x2 + 18x -16 Factors of: 9 ● -16 = -144 ◦ 9x2 +18x – 16 ◦ First, place the first 9x 9x 12 ● -12 4 ● -38 term on both sides of 3 ● -48 the bars. (without the 6 ● -24 -6 ● 24 exponent) -4 ● 36 -2 ● 72 ◦ Second, multiply the Etc…… first (9) and the last number (-16). ◦ Find the factors. Factoring Trinomials ax2 + bx + c Factoring trinomials in 9x2 + 18x -16 the form of: ax2 + bx + c. Factors of: -144 Factor: 9x2 + 18x –16 9x 9x 12 + -12 = 0 ◦ After finding the 4 + -38 = -34 factors, choose the -6 +24 3 + -48 = -45 6 + -24 = -18 set whose difference -6 + 24 = +18 -4 + 36 = +32 equals the middle -2 + 72 = +70 Etc…… number, +18. ◦ Place that set with signs under the bar. Factoring Trinomials ax2 + bx + c Factor: 9x2 + 18x –16 9x2 + 18x -16 To finish factoring you Factors of: -144 must reduce each side of the bar if possible. 9x 9x 12 + -12 = 0 4 + -38 = -34 The 9x over –6 will -6 +24 3 + -48 = -45 6 + -24 = -18 reduce to 3x over -2. 3x 3x -6 + 24 = +18 -4 + 38 = +34 The 9x over +24 will -2 +8 -2 + 72 = +70 Etc…… reduce to 3x over +8. Now, move each side into (3x -2) (3x +8) = 9x2 +18x –16 the factored answer. Factoring Trinomials ax2 + bx + c The last example contains a ‘Greatest Common Factor’: 2t2 + 3t - 5 ◦ 6t3 + 9t2 – 15t ◦ First factor out the Greatest Common Factor (GCF) which is 3t. ◦ Go ahead and place the GCF in your answer so you do not forget it. ◦ Now, place the remaining 3t( )( ) = 6t3 + 9t2 – 15t trinomial in your box with the bars. Factoring Trinomials ax2 + bx + c The last example contains a ‘Greatest Common Factor: 2t2 + 3t - 5 Factors of: ◦ 6t3 + 9t2 – 15t = 3t(2t2 +3t – -10 = 2 ● -5 5) 2t 2t 1 + -10 = -9 ◦ Next, place the first term 2 + -5 = -3 on both sides of the bars. 5 -2 5 + -2 = +3 (without the exponent) 10 + -1 = +9 ◦ Then, multiply the first number (2) and the last number (-5). ◦ Find the factors that the 3t( )( ) = 6t3 + 9t2 – 15t difference (-) will give you the middle term (+3). Place those factors under the bar. Factoring Trinomials ax2 + bx + c To finish factoring you 2t2 + 3t - 5 must reduce each side of Factors of: -10 = 2 ● -5 the bar if possible. 2t 2t 1 + -10 = -9 The left side of the bar 2 + -5 = -3 5 -2 does not reduce, but the 5 + -2 = +3 10 + -1 = +9 right side does reduce to 1t -1 1t (or t) over –1. Now, move each side into 3t(2t +5)(t - 1) = 6t3 + 9t2 – 15t the factored answer. Factoring Trinomials Tucker Method Factor these trinomials completely: 1. 6y2 + 13y + 6 2. 15t2 + 19t – 10 3. 18x2 – 24x + 6 4. 12x2 – 28x – 24 5. 15x3 + 19x2 – 10x Hand the answers into your instructor in class. Zero Product Property Let A and B be real numbers or algebraic expressions. If AB=0, then A=0 or B=0. This means that If the product of 2 factors is zero, then at least one of the 2 factors had to be zero itself! Ex: If xy=0, then either x=0 or y=0. Example: Solve. 2t2-17t+45=3t-5 2t2-17t+45=3t-5 Set eqn. =0 2t2-20t+50=0 factor out GCF of 2 2(t2-10t+25)=0 divide by 2 t2-10t+25=0 factor left side (t-5)2=0 set factors =0 t-5=0 solve for t +5 +5 t=5 check your solution! Example: Solve. x2+3x-18=0 x2+3x-18=0 Factor the left side (x+6)(x-3)=0 set each factor =0 x+6=0 OR x-3=0 solve each eqn. -6 -6 +3 +3 x=-6 OR x=3 check your solutions! Example: Solve. 3x-6=x2-10 3x-6=x2-10 Set = 0 0=x2-3x-4 Factor the right side 0=(x-4)(x+1) Set each factor =0 x-4=0 OR x+1=0 Solve each eqn. +4 +4 -1 -1 x=4 OR x=-1 Check your solutions! Finding the Zeros of an Equation The Zeros of an equation are the x- intercepts ! First, change y to a zero. Now, solve for x. The solutions will be the zeros of the equation. Example: Find the Zeros of y=x2-x-6 y=x2-x-6 Change y to 0 0=x2-x-6 Factor the right side 0=(x-3)(x+2) Set factors =0 x-3=0 OR x+2=0 Solve each equation +3 +3 -2 -2 x=3 OR x=-2 Check your solutions! If you were to graph the eqn., the graph would cross the x-axis at (-2,0) and (3,0). Assignment