Factoring,Trinomials,Quadratic education Factoring Trinomials (Quadratics) : Lucid Explanation of the

Some forms of parent involvement with the school such as communications with school, volunteering, attending school events and parent--parent connections appeared to have little effect on student achievement, especially in high school. Helpi Translation jobs are undertaken by professional translators who are well versed with at least two languages.Translation can work at two levels: inter-state or regional language translation and inter-national or foreign language translation.

Consider the product of the two linear expressions (y+a) and (y+b).(y+a)(y+b) = y(y+b) + a(y+b) = y^2 + by + ay + ab = y^2 + y(a+b) + abWe can write it asy^2 + y(a+b) + ab = (y+a)(y+b) .......(i)Similarly, Consider the product of the two linear expressions (ay+b) and (cy+d).(ay+b)(cy+d) = ay(cy+d) + b(cy+d) = acy^2 + ady + bcy + bd = acy^2 + y(ad+bc) + bdWe can write it asacy^2 + y(ad+bc) + bd = (ay+b)(cy+d) .......(ii)Equation (i) is Simple Quadratic Polynomial expressed as Product of Two linear Factorsand Equation (ii) is General Quadratic Polynomial expressed as Product of Two linear FactorsObserving the two Formulas, leads us to the method of Factorization of Quadratic Expressions.In Equation (i),the product of coefficient of y^2 and the constant term = aband the coefficient of y = a+b = sum of the factors of abSimilarly, In Equation (ii),the product of coefficient of y^2 and the constant term = (ac)(bd) = (ad)(bc)and the coefficient of y = (ad+bc) = sum of the factors of acbdSo, if we can resolve the product of y^2 and the constant term into product of two factors in such a way that their sum is equal to the coefficient of y, then we can factorize the quadratic expression.We discuss the steps involved in the method and apply it to solve a number of problems.Method of Factoring Trinomials (Quadratics) :Step 1 :Multiply the coefficient of y^2 by the constant term.Step 2 : Resolve this product into two factors such that their sum is the coefficient of yStep 3 : Rewrite the y term as the sum of two terms with these factors as coefficients.Step 4 : Then take the common factor in the first two terms and the last two terms.Step 5 : Then take the common factor from the two terms thus formed.What you get in step 5 is the product of the required two factors.The method will be clear by the following Solved Examples.The examples are so chosen that all the models are covered.Example 1 :Factorize 9y^2 + 26y + 16Solution : Let P = 9y^2 + 26y + 16Now, follow the five steps listed above.Step 1:(Coefficient of y^2) x (constant term) = 9 x 16 = 144Step 2:We have to express 144 as two factors whose sum = coefficient of x = 26;144 = 2 x 72 = 2 x 2 x 36 = 2 x 2 x 2 x 18 = 8 x 18; (8 + 18 = 26)Step 3:P = 9y^2 + 26y + 16 = 9y^2 + 8y + 18y + 16Step 4:P = y(9y + 8) + 2(9y + 8)Step 5:P = (9y + 8)(y + 2)Thus, 9y^2 + 26y + 16 = (9y + 8)(y + 2) Ans.Example 2 :Factorize y^2 + 7y - 78Solution : Let P = y^2 + 7y - 78Now, follow the five steps listed above.Step 1:(Coefficient of y^2) x (constant term) = 1 x -78 = -78Step 2:We have to express -78 as two factors whose sum = coefficient of y = 7 ;-78 = -2 x 39 = -2 x 3 x 13 = -6 x 13; (-6 + 13 = 7)Step 3:P = y^2 + 7y - 78 = y^2 - 6y + 13y - 78Step 4:P = y(y - 6) + 13(y - 6)Step 5:P = (y - 6)(y + 13)Thus, y^2 + 7y - 78 = (y - 6)(y + 13) Ans.Example 3 :Factorize 4y^2 - 5y + 1Solution : Let P = 4y^2 - 5y + 1Now, follow the five steps listed above.Step 1:(Coefficient of y^2) x (constant term) = 4 x 1 = 4Step 2:We have to express 4 as two factors whose sum = coefficient of y = -5 ;4 = 4 x 1 = -4 x -1; [(-4) + (-1) = -5]Step 3:P = 4y^2 - 5y + 1 = 4y^2 - 4y - y + 1Step 4:P = 4y(y - 1) - 1(y - 1)Step 5:P = (y - 1)(4y - 1)Thus, 4y^2 - 5y + 1 = (y - 1)(4y - 1) Ans.Example 4 :Factorize 3y^2 - 17y - 20Solution : Let P = 3y^2 - 17y - 20Now, follow the five steps listed above.Step 1:Coefficient of y^2 x constant term = 3 x -20 = -60Step 2:We have to express -60 as two factors whose sum = coefficient of x = -17 ;-60 = -20 x 3; (-20 + 3 = -17)Step 3:P = 3y^2 - 17y - 20 = 3y^2 - 20y + 3y - 20Step 4:P = y(3y - 20) + 1(3y - 20)Step 5:P = (3y - 20)(y + 1)Thus, 3y^2 - 17y - 20 = (3y - 20)(y + 1) Ans.Example 5 :Factorize 2 - 5y - 18y^2Solution : Let P = 2 - 5y - 18y^2 = -18y^2 - 5y + 2Now, follow the five steps listed above.Step 1:(Coefficient of y^2) x (constant term) = -18 x 2 = -36Step 2:We have to express -36 as two factors whose sum = coefficient of y = -5 ;-36 = -2 x 18 = -2 x 2 x 9 = 4 x -9; [4 + (-9) = -5]Step 3:P = -18y^2 - 5y + 2 = -18y^2 + 4y - 9y + 2Step 4:P = 2y(-9y + 2) + 1(-9y + 2)Step 5:P = (-9y + 2)(2y + 1)Thus, 2 - 5y - 18y^2 = (-9y + 2)(2y + 1) Ans.Example 6 :Factorize (y^2 + y)^2 -18(y^2 + y) + 72Solution : Let P = (y^2 + y)^2 -18(y^2 + y) + 72Put (y^2 + y) = t; Then P = t^2 -18t + 72Now, follow the five steps listed above.Step 1:(Coefficient of t^2) x (constant term) = 1 x 72 = 72Step 2:We have to express 72 as two factors whose sum = coefficient of t = -18 ;72 = 12 x 6 = -12 x -6; [(-12) + (-6) = -18]Step 3:P = t^2 -18t + 72 = t^2 - 12t - 6t + 72Step 4:P = t(t - 12) - 6(t - 12)Step 5:P = (t - 12)(t - 6)But t = (y^2 + y);So, P = (t - 12)(t - 6) = (y^2 + y - 12)(y^2 + y - 6)In each of these two brackets, there is a Quadratic Polynomial which can be factorised using the five steps above.y^2 + y - 12 = y^2 + 4y - 3y - 12 = y(y + 4) - 3(y + 4) = (y + 4)(y - 3)y^2 + y - 6 = y^2 + 3y - 2y - 6 = y(y + 3) - 2(y + 3) = (y + 3)(y - 2)See how these two Quadratic Polynomials are factorised with the knowledge of the 5 steps.You might have mastered the 5 steps of factorisation by this time, to write directly like this.Thus,P = (y^2 + y)^2 -18(y^2 + y) + 72= (y^2 + y - 12)(y^2 + y - 6)= (y + 4)(y - 3)(y + 3)(y - 2) Ans.For more, on Factoring Quadratics, go to http://www.math-help-ace.com/Factoring-Trinomials.html Article Tags: Factoring Trinomials Quadratics, Five Steps Listed, Steps Listed Above, Listed Above Step, Factoring Trinomials, Trinomials Quadratics, Quadratic Polynomial, Constant Term, Five Steps, Steps Listed, Listed Above, Above Step, Factors Whose, Y)^2 -18(y^2