# Identify the case that gives 7 x 2 x 2 7 x 15 2 x 3 x

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Identify the case that gives7x.2x27x15 = (2x+ 3)(x5)You can check the answer to Example 9 by multiplying out the brackets.The solution to Example 9 includes some working to help with finding thefactorisation. You might find it helpful to write down working like thiswhile you’re getting used to factorising quadratics, but after a while you’llprobably find that you can usually do it in your head. So, when youfactorise a quadratic, you don’t need to write down any working – it’s fineto just write down the quadratic and its factorisation, like this:2x27x15 = (2x+ 3)(x5).Here’s a summary of the method demonstrated in the example above.
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Unit 2Graphs and equationsStrategy:To factorise a quadratic of the formax2+bx+c1.Take out any numerical common factors. If the coeﬃcient ofx2isnegative, also take out the factor1. Then apply the steps belowto the quadratic inside the brackets.2.Find the positive factor pairs ofa, the coeﬃcient ofx2. For eachsuch factor paird, ewrite down a framework (dx)(ex).3.Find all the factor pairs ofc, the constant term (including bothpositive and negative factors).4.For each framework and each factor pair ofc, write the factorpair in the gaps in the framework in both possible ways.5.For each of the resulting cases, calculate the term inxthat youobtain when you multiply out the brackets.6.Identify the case where this term isbx, if there is such a case.This is the required factorisation.As with the earlier strategy, if this strategy doesn’t lead to a factorisation,then the quadratic can’t be factorised using integers.Activity 18Factorising quadratics of the formax2+bx+cFactorise the following quadratics. (Theycanall be factorised.)(a) 5x2+ 13x6(b) 3x2+ 16x+ 5(c) 6x211x+ 3(d) 5x28x21(e) 18x2+ 9x2(f) 4x28x+ 3(g) 4p219p5(h) 6u2+ 11u35(i) 4t2+ 4t+ 1(j) 9v212v+ 4(k)4s2+ 4s+ 3(l) 12y210y2There are two special types of quadratic that can be factorised more easilythan those that you have seen so far in this subsection. You should alwayscheck whether your quadratic is one of these before you embark on eitherof the two strategies above.Quadratics with no constant termThese can be factorised by takingxout as a common factor. For example,x2+ 4x=x(x+ 4),3x26x= 3(x22x) = 3x(x2).162
4QuadraticsDifferences of two squaresAs you saw in Unit 1, adifference of two squaresis any expression ofthe formA2B2,whereAandBare subexpressions. You can check, by multiplying out thebrackets, thatA2B2= (A+B)(AB).(19)If you can recognise a quadratic as a difference of two squares, then youcan use equation (19) to factorise it immediately. For example,x29 =x232= (x+ 3)(x3),x21 =x212= (x+ 1)(x1),4x21 = (2x)212= (2x+ 1)(2x1).

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Term
Summer
Professor
NoProfessor
Tags
Euclidean geometry, Line segment, Vincenzo Riccati
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