Presentation on theme: "Factorisation of Binomials, Trinomials, Sum & Difference of Two Cubics"— Presentation transcript:
1Factorisation of Binomials, Trinomials, Sum & Difference of Two Cubics Revision Algebra IIFactorisation of Binomials, Trinomials, Sum & Difference of Two CubicsBy I Porter
2FactorsA factor of a term or number divides into that term or number, without a remainder.For examples, the factors of18 are 1, 2, 3, 6, 9 and 18.6x2 are 1, 2, 3, 6, x, 2x, 3x, 6x, x2, 2x2, 3x2 and 6x2.4xy are 1, 2, 4, x, y, 2x, 2y, 4x, 4y, xy, 2xy and 4xy.The highest common factor (HCF) of 4xy and 6x2 is 2x.The product of a(b + c) is ab + ac and the product of (a + b)(c + d) is ac + ad + bc + bd, we say, these two results are written in expanded form.When they are written in the form a(b + c) and (a + b)(c + d), we say they are in factored form.Factorisation is the process whereby an expression in expanded form, such as ab + ac,is changed to factored form, such as a(b + c).
3Factorising by Grouping Algebraic expressions with 4 terms do not have a factor common to every term.Such expressions can often be factorised by first grouping the terms in pairs.= x(a + b)+ y(a+ b)= (a + b)ax+ bx + ay +by(x + y)Examples: Fully factorise the followinga) a(b - 3) + 6(b - 3)b) ac + ad + bc + bdc) 4x2 - 4xy + x - y= a(b - 3) + 6(b - 3)= ac + ad + bc + bd= 4x2 - 4xy + x - y= (b - 3)(a + 6)= a(c + d)+ b(c + d)= 4x(x - y)+ 1(x - y)= (c + d)(a + b)= (x - y)(4x + 1)
7Factoring quadratic trinomials: ax2 ± bx ± c A trinomial has 3 terms. Exercise: Factorisea) xb) 8x3 + y3c) 9x3 - 72y3= (x - 5)(x2 + 5x + 25)= (2x + y)(4x2 - 2xy + y2)= 9(x - 2y)(x2 + 2xy + 2y2)Factoring quadratic trinomials: ax2 ± bx ± cA trinomial has 3 terms.A quadratic trinomial is an expression with 3 terms where the highest power ofa term is 2.Sum and Product Method.To factorise a trinomial of the form x2 ± bx ± c (where b and c are non-zero integers) look for two numbers P and Q such that:* Their sum is b : (P + Q = b)* Their product is c : (PQ = c)[ The first step to start may be to write out the factors of c in a systematic process.]
8Examples: Factorise the following: a) x2 + 7x +12Sum = +2Product = -24Sum = +7Product = +12c) x2 + 2x - 24Factors of -24-24 = -1 x 24= -2 x 12= -3 x 8= -4 x 6Factor sums= +23= +10= +5= +2Factors of +1212 = 1 x 12= 2 x 6= 3 x 4Factor sums= +132 + 6 = +83 + 4 = +7Why not the (-) factors?Sum and Product are both (+)Why do the factors have different signs?Sum (+) and Product (-), largest number is (+)So, x2 + 7x +12 = (x + 3)(x + 4)So, x2 + 2x - 24 = (x - 4)(x + 6)Sum = -3Product = -40b) x2 - 8x +12Sum = -8Product = +12d) x2 - 3x - 40Factors of -40-40 = 1 x -40= 2 x -20= 4 x -10= 5 x -8Factor sums= -39= -18= -6= -3Factors of +1212 = -1 x -12= -2 x -6= -3 x -4Factor sums= -13= -8= -7Why not the (+) factors?Sum (-) and Product (+) both are (-)Why do the factors have different signs?Sum (-) and Product (-), largest number is (-)So, x2 - 8x +12 = (x - 2)(x - 6)So, x2 - 3x - 40 = (x + 5)(x - 8)
10Harder factorisation of quadratic trinomials: ax2 + bx + c Exercise: Factorisea) 7x2 - 63x + 140b) 12x2 - 12x - 72c) 9x2 + 36x + 27= 7(x - 4)(x - 5)= 12(x + 2)(x - 3)= 9(x + 1)(x + 3)Harder factorisation of quadratic trinomials: ax2 + bx + cExamples: Factorisea) 3x2 - 2x - 8b) 6x2 + 13x - 5[ Variation of sum + product method]Step 1: multipy first and last term. 3 x -8 = -24Step 2: Look for two numbers that have a sum of -2 and product of -24. (I.e. -6 and +4)Step 3: Split the middle term using the number found in step 23x2 - 6x + 4x - 8Step 4: Factorise.Step 1: multipy first and last term. 6 x -5 = -30Step 2: Look for two numbers that have a sum of +13 and product of -30. (I.e. -2 and +15)Step 3: Split the middle term using the number found in step 26x2 - 2x + 15x - 5Step 4: Factorise.3x2 - 2x - 8= 3x2 - 6x + 4x - 86x2 + 13x - 5= 6x2 - 2x + 15x - 5= 3x(x - 2) + 4(x - 2)= 2x(3x - 1) + 5(3x - 1)= (x - 2) (3x + 4)= (3x - 1) (2x + 5)If you can use the CROSS method successfully, KEEP USING IT!