Mathematics 506 - Technical & Scientific Option Lesson 2 β Factoring Polynomials (contβd) Methods: 5) Perfect Square Trinomials 6) Completing the Square 7) Roots Method Perfect Square Trinomials A trinomial is a perfect square when the middle term is equal to twice the product of the square roots of the end terms. π2 + 2ππ + π 2 = (π + π)2 Ex. OR 9π₯ 2 + 12π₯ + 4 π2 = 9π₯ 2 βΉ π = 3π₯ π2 = 4 βΉ π=2 βΉ 2ππ = 2 3π₯ 2 = 12π₯ β΄ 9π₯ 2 + 12π₯ + 4 = (3π₯ + 2)2 3π₯ + 2 3π₯ + 2 Verify: = 9π₯ 2 + 6π₯ + 6π₯ + 4 = 9π₯ 2 + 12π₯ + 4 Ex. 9π₯ 2 β 12π₯ + 4 π2 = 9π₯ 2 βΉ π = 3π₯ π2 = 4 βΉ π=2 βΉ 2ππ = 2 3π₯ 2 = 12π₯ β΄ 9π₯ 2 β 12π₯ + 4 = (3π₯ β 2)2 π2 β 2ππ + π 2 = (π β π)2 Mathematics 506 - Technical & Scientific Option Completing the Square If the trinomial is not a perfect square we can complete the square by adding a constant. NB: This method only works when the coefficient of π₯ 2 is 1. If it is not, the expression must be factored (or the equation must be divided by the coefficient) before we can complete the square. Ex. 2π₯ 2 β 12π₯ + 8 = 2(π₯ 2 β 6π₯ + 4) To complete the square π₯ 2 + ππ₯, add the square of half the coefficient of π₯ (ie. π 2 ) 2 to create a perfect square. We must also subtract this same constant from the expression. π₯ 2 + ππ₯ + Ex. π 2 2 β π 2 2 π + π = (π₯ + 2 )2 β π 2 2 + π5 2π₯ 2 β 12π₯ + 8 = 2(π₯ 2 β 6π₯ + 4) Step 1: Make the coefficient of π₯ 2 =1. Step 2: Find π 2 2 = β6 2 2 = 2[(π₯ 2 β 6π₯ + 9) β 9 + 4] Step 3: Add (and subtract ) this constant. = 2[(π₯ β 3)2 β 5] Step 4: Factor the perfect square. = 2(π₯ β 3)2 β 10 Step 5: Simplify This method is used mainly when solving equations. Ex. =9 2π₯ 2 β 12π₯ + 8 = 0 2(π₯ β 3)2 β 10 = 0 Factor as explained above 2(π₯ β 3)2 = 10 Solve for π₯ (π₯ β 3)2 = 5 π₯β3=± 5 Case 1: π₯ β 3 = 5 π₯ =3+ 5 Case 2: π₯β3=β 5 π₯ =3β 5 Mathematics 506 - Technical & Scientific Option Roots Method Any second degree trinomial (ππ₯ 2 + ππ₯ + π) can be factored if and only if the discriminant ο is positive or zero, where ο= π 2 β 4ππ If ο β₯ 0 then ππ₯ 2 + ππ₯ + π = π π₯ β π₯1 π₯ β π₯2 and π₯1 = βπ+ β³ 2π = βπ+ π 2 β4ππ 2π This method is also mainly used to solve equations. π₯2 = βπβ β³ 2π = βπβ π 2 β4ππ 2π
© Copyright 2026 Paperzz