Factorising Harder

QuadraticsSlideshow 16 　

MathematicsMr Sasaki　　　 Room

307

Objectives

•Factorise quadratics in the form where and are divisible by .•Factorise other quadratics in the form

Easy QuadraticsWe can use methods we already know to solve some quadratics in the form .ExampleSolve .

¿2(𝑥−3)(𝑥+2)If we can divide all terms by the coefficient, we can remove it and factorise the quadratic in the form .

Answers2(𝑥+1)(𝑥+2) 2(𝑥+3)(𝑥+5)3 (𝑥+2)(𝑥+1) 2(𝑥−1)(𝑥+3)2(𝑥−3)(𝑥+4 ) 4(𝑥+1)(𝑥+2)3 (𝑥+5)(𝑥−7) 4(𝑥+7)(𝑥−2)3 (𝑥−5)(𝑥−3) 5(𝑥+3)(𝑥−2)7 (𝑥−3)(𝑥−1) 2(𝑥+5)(𝑥−7)3 (𝑥−11)(𝑥+2) 4(𝑥+12)(𝑥−4)2(𝑥−7)(𝑥+16) 6 (𝑥−13)(𝑥+31)

Typical QuadraticsAs you should know, we were lucky with the questions on the last worksheet. Usually it’s more complicated.Let’s think back to expressions in the form .What are our two simultaneous equations for the numbers and ?

𝑎=𝑐+𝑑 𝑏=𝑐 ∙𝑑We also use this idea for quadratics in the form .

Typical QuadraticsFor a quadratic in the form , we need to consider two numbers.ExampleFactorise.

×1𝑎 ∙𝑏=¿ 𝑎+𝑏=¿

(Or vice-versa.)

We can write the coefficient of as the sum of .

2 𝑥2+𝑥−6=¿2 𝑥2−3 𝑥+4 𝑥−6Finally, we group the first two and last two terms.(2 𝑥2−3 𝑥)+(4 𝑥−6 )¿𝑥 (2𝑥−3)+¿2(2𝑥−3)

¿(𝑥+2)(2𝑥−3)

Typical QuadraticsLet’s try one more example.ExampleFactorise.

×−48 8𝑎 ∙𝑏=¿ 𝑎+𝑏=¿

(Or vice-versa.)

6 𝑥2+8 𝑥−8=¿6 𝑥2−4 𝑥+12𝑥−8¿ (6 𝑥2−4 𝑥)+(12𝑥−8)¿2 𝑥(3𝑥−2)+¿4(3𝑥−2)

¿ (2𝑥+4)(3 𝑥−2)¿2(𝑥+2)(3 𝑥−2)

Answers(2 𝑥+1)(𝑥+3) (3 𝑥+2)(𝑥−1)(2 𝑥−3)(𝑥+1) (4 𝑥−1)(𝑥+2)(2 𝑥+1)(2 𝑥+3) (4 𝑥+3)(2𝑥+3)(2 𝑥−1)(3 𝑥+2) (4 𝑥+5)(2𝑥+7)(𝑥+2)(2𝑥−7) 2(𝑥−2)(2𝑥+1)(5 𝑥−2)(2𝑥+5) (2 𝑥−1)(2 𝑥+1)2(𝑥+2)(2 𝑥−1) 2(3𝑥+2)(𝑥−4)6 (4 𝑥−1)(𝑥+1) 3 (2𝑥+7)(2𝑥−7)(5 𝑥+4 )2 (8 𝑥+1 )2

3 (2𝑥−5)(𝑥−9) 2(2𝑥−9)(3 𝑥+5)(3 𝑥−2)(7𝑥+1) (5 𝑥+8)(3 𝑥−11)−2 (3 𝑥+2)(2𝑥+5)2 𝑥(3𝑥+4)(2 𝑥+3)

Squaring / Square RootingLet’s have a bit of practice!16 49 144±3 ±7 ±9196 289 256±8 ±11 ±11 400 529±12 ±14 ±18729 576 1296±21 ±30 ±272304 2809 3844± 43 ± 48 ±528281 10816 16129±67 ±72 ±89

Evil QuadraticsSome of those quadratics were tough to factorise because some big numbers have a lot of factors…or factors that are hard to find.

For a difficult quadratic, it may be more sensible to pretend its equal to , solve it and then factorise the expression.If a quadratic equation has solutions or , how does it factorise?, we learned this from Grade 8 last year! This is correct…right?

Evil QuadraticsDo you remember the formula for solving a quadratic equation?For some 𝑥=−𝑏±√𝑏2−4𝑎𝑐

2𝑎Let’s try factorising using this with an easy example.ExampleFactorise .First, let . Now…

𝑥=−𝑏±√𝑏2−4𝑎𝑐2𝑎 =¿

−3±√32−4 ∙1∙22 ∙1

=¿−3±√12

We get . So how does our expression factorise?(𝑥+1)(𝑥+2)

Now some of those quadratics on the hard sheet were tough. Let’s try again.Example Factorise .Let . (

𝑥=−𝑏±√𝑏2−4𝑎𝑐2𝑎 =¿

31±√(−31)2−4 ∙15 ∙(−88)2∙15

¿ 31±√961+528030 ¿ 31±√6 241

30¿31±7 930

We get . So we have… (𝑥− 113 )(𝑥+ 85 )

Is this correct?No! This is factorised.We need to multiply it by .

Evil Quadratics

In fact, when we solve and get two solutions and …we substitute these into…

𝑎(𝑥−𝑥1)(𝑥−𝑥2)So, as , and , we get…

15(𝑥− 113 )(𝑥+ 85 )¿3 (𝑥− 113 )∙5(𝑥+ 8

5 )¿ (3 𝑥−11) (5𝑥+8 )

And that’s it! It is a long process but it’s easier with big numbers, especially when is a big product! Let’s try some questions using this method.

Evil Quadratics

Answers - Regular

2(3𝑥+5)(𝑥−3) 3 (2𝑥−5)(𝑥+3)

(4 𝑥−1)(2 𝑥−11) (5 𝑥+2)(2𝑥−5)

3 (𝑥+3)(𝑥+4) (2 𝑥+3)(2𝑥−9)

2(2𝑥+3)(5 𝑥−7 )2(𝑥+4)(𝑥+21)

Answers - Applied

(any order)

𝑎=− 14 ,𝑏=13

mm

(𝑎+𝑏)(𝑎−𝑏)

12𝑎𝑛𝑑13

2𝑚