Quadratic Algebra Lesson #2

Transcription

1 Quadratic Algebra Lesson # Factorisation Of Quadratic Expressions Many of the previous expansions have resulted in expressions of the form ax + bx + c. Examples: x + 5x+6 4x 9 9x + 6x + 1 These are known as quadratic expressions. A quadratic expression in x is an expression of the form ax + bx + c where x is the variable and a, b and c represent constants with a 0. Note: If a = 0, then the expression becomes bx + c which is linear in x. In the following sections you will learn several techniques for factorising quadratic expressions and solving quadratic equations. Factorisation is critical in the solution of problems that convert to quadratic equations. Factorisation is the process of writing an expression as a product of factors. For example: given the product of two factors such as (x + )(x + 3), we can expand to get (x + )(x + 3) = x + 3x + x + 6 = x + 5x + 6. Notice that: Factorisation is the reverse process of expansion. For example: As x + 5x + 6 = (x + )(x + 3), we say that (x + ) and are factors of x + 5x+6. (x + 3)

2 So far in this section of the course, we have used the distributive law to expand binomial products of the form (a + b)(c + d), i.e., (a + b)(c + d) = ac + ad + bc + bd, using the FOIL method. Special expansions you should be familiar with include: (x + p)(x + q) = x + (p + q)x + pq {sum and product expansion} (x + a) = x + ax + a {perfect square expansion} (x a) = x ax + a (x + a)(x a) = x a {difference of two squares expansion} These statements are called identities because they are true for all values of the variable x. Notice that the RHS of each identity is a quadratic expression which has been formed by expanding the LHS. The LHS of the identities above can be obtained by factorising the RHS. Removal Of Common Factors Many quadratic expressions are easily factorised by removing the Highest Common Factor (HCF). In fact, we should always look to remove the HCF before proceeding with any other factorisation. Example 1: Factorise by removing a common factor: (a.) x + 3x (b.) x 6x (a.) x + 3x has HCF x (b.) x 6x has HCF x = x(x + 3) = x(x 3) Example : Fully factorise by removing a common factor: (a.) (x 5) (x 5) (b.) (x + ) + x + 4 (a.) (x 5) (x 5) = (x 5)(x 5) (x 5) has HCF (x 5) = (x 5)[(x 5) ] = (x 5)(x 7) {simplifying}

3 (b.) (x + ) + x + 4 = (x + )(x + ) + (x + ) has HCF (x + ) = (x + )[(x + ) + ] = (x + )(x + 4) The Difference Of Two Squares Factorisation We know that the expansion of (a + b)(a b) is a b. Therefore, the factorisation of a b is (a + b)(a b), i.e., a b = (a + b)(a b). Example 1: Use the rule a b = (a + b)(a b) to factorise fully: (a.) 9 x (b.) 4x 5 (a.) 9 x (b.) 4x 5 = 3 x = (x) 5 = (3 + x)(3 x) = (x + 5)(x 5) Example : Fully factorise: (a.) x 8 (b.) 3x 48 (a.) x 8 = (x 4) {HCF is } = (x ) {difference of squares} = (x + )(x ) (b.) 3x 48 = 3(x 16) {HCF is 3} = 3(x 4 ) {difference of squares} = 3(x + 4)(x 4) We notice that x 9 is the difference between two squares and therefore, we can factorise it using a b = (a + b)(a b). Even though 7 is not a perfect square, we can still factorise x 7 if we use 7 = ( 7). o So, x 7 = x ( 7) = (x + 7)(x 7) We say that x+ 7 and x 7 are the linear factors of x 7.

4 Example 3: Factorise into linear factors: (a.) x 11(b.) (x + 3) 5 (a.) (b.) x 11 = x ( 11) (x + 11)(x 11) (x + 3) 5 = (x + 3) ( 5) = [(x + 3) + 5][(x + 3) 5] [x ][x + 3 5] Example 4: Factorise using the difference between two squares: (a.) (3x + ) 9 (b.) (x + ) (x 1) (a.) (b.) (3x + ) 9 = (3x + ) 3 = [(3x + ) + 3][(3x + ) 3] = [3x + 5][3x 1] (x + ) (x 1) = [(x + ) + (x 1)][(x + ) (x 1)] = [x + + x 1][x + x 1] = [x + 1][3] = 3(x + 1) Perfect Square Factorisation We know the expansion of (x + a) is x + ax+ a and (x a) is x ax+ a. Therefore, the factorisation of x + ax+ a is (x + a) and x ax+ a is (x a), o i.e., x + ax+ a = (x+ a) o x ax+ a = (x a)

5 Example 1: Use perfect square rules to fully factorise: (a.) x + 10x+ 5 (b.) x 14x+ 49 (a.) x + 10x+ 5 (b.) x 14x+ 49 = x + x 5+ 5 = x x 7+ 7 = (x + 5) = (x 7) Example : Fully factorise: (a.) 9x 6x + 1 (b.) 8x + 4x + 18 (a.) (b.) 9x 6x + 1 = (3x) 3x 1+ 1 = (3x 1) 8x + 4x + 18 = (4x + 1x + 9) = ([x] + x ) = (x + 3) {as HCF is } Quadratic Trinomial Factorisation A quadratic trinomial expression has an term and three terms overall. For example: x + 7x+6 and 3x 13x 10. Notice that (x+ p)(x+ q) = x + qx+ px+ pq= x + (p+ q)x+ pq, so: x + qx+ px+ pq= x + (p+ q)x+ pq = (x+ p)(x+ q). So, in order to factorise x + 7x+ 6, we look for two numbers with a product of 6 and a sum of 7. These numbers are 1 and 6, so x + 7x + 6 = (x + 1)(x + 6). x

6 Example 1: Use the sum and product method of factorisation to fully factorise: (a.) x + 5x+4 (b.) x x 1 (a.) x + 5x+4 has p+ q= 5 and pq = 4. So, p and q are 1 and 4. Therefore, x + 5x + 4 = (x + 1)(x + 4). (b.) x x 1 has p+ q= 1 and pq = 1. So, p and q are -4 and 3. Therefore, x x 1 = (x 4)(x + 3). Example : Fully factorise by first removing a common factor. (a.) 3x 9x + 6 (b.) x x 1 (a.) (b.) 3x 9x + 6 = 3(x 3x + ) {removing 3 as a common factor} = 3(x )(x 1) {as p+ q= 3 and pq = gives p =, q= 1} x x 1 = (x x 6) {removing as a common factor} = (x 3)(x + ) {as p+ q= 1 and pq = 6 gives p = 3, q= } Miscellaneous Factorisation Use the following steps in order to factorise quadratic expressions: o Step 1: Look at the quadratic expression to be factorised. o Step : If there is a common factor, take it out. o Step 3: Look for perfect square factorisation: x + ax+ a = (x+ a) or x ax+ a = (x a) or look for the difference of two squares: x a = (x + a)(x a) or look for the sum and product type: x + (p + q)x + pq = (x + p)(x + q)

7 Factorisation of ax + bx + c (a 0) In the previous section we revised techniques for factorising quadratic expressions in the form ax + bx + c where: o a = 1 For example: o a was a common factor For example: x + 5x+6 x + 10x + 1 = (x + 3)(x + ) = (x + 5x + 6) = (x + 3)(x + ) Factorising a quadratic expression such as 3x + 11x + 6 appears to be more complicated because the coefficient of is not one and is not a common factor. We need to develop a method for factorising this type of quadratic expression. Two methods for factorising ax + bx + c where a 1 are commonly used. These are: o trial and error o by splitting the x-term Trial And Error For example: consider the quadratic 3x + 13x + 4. x Since 3 is a prime number, To fill the gaps, we are seeking two numbers with product of 4 and net result of inners and outers being 13 x. As the product is 4, we will try and, 4 and 1, and 1 and 4. o (3x+ )(x+ ) = 3x + 6x+ x+ 4 fails o (3x+ 4)(x+ 1) = 3x + 3x+ 4x+ 4 fails o (3x+ 1)(x+ 4) = 3x + 1x+ x+ 4 is successful o So, 3x + 13x + 4 = (3x + 1)(x + 4)

8 Note: We could set these trials out in table form: Now, if a and c are not prime in ax + bx + c there can be many possibilities. For example, consider 8x + x By simply using trial and error, the possible factorisations are: (8x+ 5)(x+ 3) x (4x + 5)(x + 3) this is correct (8x+ 3)(x+ 5) x (4x + 3)(x + 5) x (8x+ 1)(x+ 15) x (4x + 1)(x + 15) x (8x + 15)(x + 1) x (4x+ 15)(x+ 1) x We could set these trials out in table form: or As you can see, this process can be very tedious and time consuming. Factorisation By Splitting The x-term Using the distributive law to expand, we see that: (x + 3)(4x + 5) = 8x + 10x + 1x + 15 = 8x + x + 15 We will now reverse the process to factorise the quadratic expression 8x + x o Notice that: 8x + x + 15 = 8x + 10x + 1x + 15 {splitting the middle term} = (8x + 10x) + (1x + 15) {grouping in pairs} = x(4x + 5) + 3(4x + 5) {factorising each pair separately} = (4x + 5)(x + 3) {completing the factorisation}

9 But how do we correctly split the middle term? That is, how do we determine that x must be written as + 10x + 1x? When looking at we notice that 8 15 = 10 and 10 1 = 10 and also =. So, in 8x + x + 15, we are looking for two numbers such that their sum is and their product is 8 15 = 10 and these numbers are 10 and 1. Likewise in 6x + 19x + 15 we seek two numbers of sum 19 and product 6 15 = 90. These numbers are 10 and 9. So, 6x + 19x + 15 = 6x + 10x + 9x + 15 = (6x + 10x) + (9x + 15) = x(3x + 5) + 3(3x + 5) = (3x + 5)(x + 3) Rules for splitting the x-term: o The following procedure is recommended for factorising ax + bx + c : Step 1: Find ac and then the factors of ac which add to b. Step : If these factors are p and q, replace bx by px + qx. Step 3: Complete the factorisation. Example 1: Show how to split the middle term of the following so that factorisation can occur: (a.) 3x + 7x + (b.) 10x 3x 5 (a.) In 3x + 7x +, ac = 6 and b = 7, so we are looking for two numbers with a product of 6 and a sum of 7. These are 1 and 6. So, the split is 7x = x + 6x. (b.) In 10x 3x 5, ac = 50 and b = 3, so we are looking for two numbers with a product of 50 and a sum of 3. These are 5 and. So, the split is 3x = 5x + x.

10 Example : Factorise, using the splitting method : (a.) 6x + 19x + 10 (b.) 3x x 10 (a.) 6x + 19x + 10 has ac = 60 and b = 19, so we are looking for two numbers with a product of 60 and a sum of 19. Searching amongst the factors of 60, only 4 and 15 have a sum of 19. Therefore, 6x + 19x + 10 = 6x + 4x + 15x + 10 {splitting the x-term} = x(3x + ) + 5(3x + ) {factorising in pairs} = (3x + )(x + 5) {taking out the common factor} (b.) 3x x 10 has ac = 30 and b = 1, so we are looking for two numbers with a product of 30 and a sum of 1. Searching amongst the factors of 30, only 5 and 6 have a sum of 1. Therefore, 3x x 10 = 3x + 5x 6x 10 = x(3x + 5) (3x + 5) = (3x+ 5)(x ) {splitting the x-term} {factorising in pairs} {taking out the common factor}