Maths Level 4 Module 3

Mathematics NQF Level 4 *see terms and conditions Downloaded from hlayiso.com Mathematics NQF Level 4

Functions and algebra Topic 2 Downloaded from hlayiso.com Mathematics NQF Level 4

Work with algebraic expressions using the remainder and factor theorems Module 3 Downloaded from hlayiso.com Mathematics NQF Level 4

Synthetic division Click above to play this video. Downloaded from hlayiso.com Example 3.1

Learning activity 3.1 Downloaded from hlayiso.com Mathematics NQF Level 4

Using synthetic division, find the quotient for the following: 2 1. x − 9 x − 10 divided by x + 1 STEP 1: x +1 =0 [ Equate divisor to 0] x = -1 STEP 2: −1  Write into  divsion box  STEP 3: 1 -9 -10  Write co-efficients  −1 in descending order  Downloaded from hlayiso.com Learning activity 3.1

1. Continued STEP 4: 1 -9 -10  Bring first  −1 coefficient down  ↓ 1  Multiply brought  1 -9 -10 down number by  STEP 5: −1 ↓ -1  number in  division box  1 STEP 6: 1 -9 -10  Add numbers in  −1 second column  ↓ -1 1 -10 Downloaded from hlayiso.com Learning activity 3.1

1. Continued STEP 7: 1 -9 -10  Multiply result of  −1 addition by number  ↓ -1 10 in division box  1 -10 1 -9 -10 Continue until  STEP 8: −1  bottom row is  ↓ -1 10 complete  1 -10 0 2 x − 9 x − 10 0 STEP 9: Final answer: =x − 10 + x +1 x +1 Downloaded from hlayiso.com Learning activity 3.1

Learning activity 3.2 Downloaded from hlayiso.com Mathematics NQF Level 4

1. If f (= x) 3 x( x + 3) and g ( x)= x + 3 a) Find the quotient q(x) f ( x) r ( x) = q( x) + g ( x) g ( x) 3 x( x + 3) ∴ 3x = ( x + 3) ∴ q( x) 3 = x Downloaded from hlayiso.com Learning activity 3.2

1. If f (= x) 3 x( x + 3) and g ( x)= x + 3 b) Find the remainder r(x) r ( x) = 0 c) Write in the form f ( x) ≡ g ( x) ⋅ q( x) + r ( x). f ( x) g ( x).q ( x) + r ( x) = 3 x( x + 3) = ( x + 3)3 x + 0 Downloaded from hlayiso.com Learning activity 3.2

Learning activity 3.3 Downloaded from hlayiso.com Mathematics NQF Level 4

e)2.2Use x + 2the 3 x −remainder 2 a − 12 xa + 6theorem 2 a 3 to calculate the x + 3remainder a in each of the following cases: VIDEO Click above to play this video. Downloaded from hlayiso.com Learning activity 3.3

Learning activity 3.5 Downloaded from hlayiso.com Mathematics NQF Level 4

3 2 x + 4 x the c)Factorise − 11following x+6 third-degree polynomials: VIDEO Click above to play this video. Downloaded from hlayiso.com Learning activity 3.5

Summative assessment Module 3 Downloaded from hlayiso.com Mathematics NQF Level 4

1. Use the remainder theorem to calculate the remainder 3 2 1.1) 2 x + 4 x − 8 x − 2 x−2 x−2=0 ∴x =2 3 2 f ( x) = 2 x + 4 x − 8 x − 2 f (2) remainder = 2(2)3 + 4(2) 2 − 8(2) − 2 [Remainder = 16 + 16 − 16 − 2 theorem] Downloaded= 14from hlayiso.com Summative assessment – Module 3

1. Use the remainder theorem to calculate the remainder 2 3 1.2) 2 − 6 x + 4 x − x 2x − 4 2x − 4 =0 ∴x = 2 [Remainder 2 3 theorem] f ( x) =2 − 6 x + 4 x − x 2 3 f = (2) remainder = 2 − 6(2) + 4(2) − 2 =2 − 12 + 16 − 8 = −2 from hlayiso.com Downloaded Summative assessment – Module 3

2. Prove that x − 2 is a factor of 2 x − 7 x + 12 x − 12 3 2 3 2 f ( x) = 2 x − 7 x + 12 x − 12 f (2) = 2(2)3 − 7(2) 2 + 12(2) − 12 [Factor = 16 − 28 + 24 − 12 theorem] =0 ∴ x − 2 is a factor Downloaded from hlayiso.com Summative assessment – Module 3

3. Test whether x + 1 is a factor of 3 2 2 x + 5 x − 9 x − 12 3 2 f ( x) = 2 x + 5 x − 9 x − 12 3 2 f (−1) = 2(−1) + 5(−1) − 9(−1) − 12 =−2 + 5 + 9 − 12 [Factor =0 theorem] ∴ x + 1 is a factor Downloaded from hlayiso.com Summative assessment – Module 3

4. Calculate the value of a when 3 2 4 x − 2 x − ax + 12 gives a remainder of 18. x+3 f ( x) = 4 x 3 − 2 x 2 − ax + 12 f (−3) = 18 ∴ 4(−3)3 − 2(−3) 2 − a (−3) + 12 = 18 ∴−108 − 18 + 3a + 12 = 18 ∴ 3a = 132 ∴a =44 Downloaded from hlayiso.com Summative assessment – Module 3

5. Factorise the following polynomials: 5.1) x3 + 125 3 f ( x) = x + 125 f (−5) =(−5)3 + 125 =0 ∴ x + 5 is a factor Downloaded from hlayiso.com Summative assessment – Module 3

5. Factorise the following polynomials: 5.1) x3 + 125 (Continued) x 2 −5x +25 x + 5 x 3 + 0 x 2 + 0 x + 125 3 2 x + 5x 2  − 5 x + 0 x + 125 −5 x 2 − 25 x  25 x + 125 25 x + 125   Downloaded from hlayiso.com Summative assessment – Module 3

5. Factorise the following polynomials: 5.1) x3 + 125 (Continued) ∴ x 3 + 125 = ( x + 5)( x 2 − 5 x + 25) Downloaded from hlayiso.com Summative assessment – Module 3

5. Factorise the following polynomials: 5.2) 2 x 3 − x 2 − 15 x + 18 Select some x-values to try to find a factor: 3 2 f ( x) = 2 x − x − 15 x + 18 f (1)= 2(1)3 − 12 − 15(1) + 18 We want to find the = 2 − 1 − 15 + 18 x-value that will make =4 f(x) = 0 ≠0 ∴ x − 1 is not a factor Downloaded from hlayiso.com Summative assessment – Module 3

5. Factorise the following polynomials: 5.2) 2 x 3 − x 2 − 15 x + 18 Continued f (−1) = 2(−1)3 − (−1) 2 − 15(−1) + 18 =−2 − 1 + 15 + 18 = 30 ≠0 ∴ x + 1 is not a factor 3 2 f (2)= 2(2) − 2 − 15(2) + 18 = 16 − 4 − 30 + 18 = 0 ∴ x − 2 is a factor Downloaded from hlayiso.com Summative assessment – Module 3

5. Factorise the following polynomials: 5.2) 2 x 3 − x 2 − 15 x + 18 Continued Divide the polynomial by the factor: 2x 2 +3x −9 x − 2 2 x3 − x 2 − 15 x + 18 Use long division or 3 2 2x − 4x synthetic division  3 x 2 − 15 x + 18 3x 2 − 6 x  − 9 x + 18 −9 x + 18   Downloaded from hlayiso.com Summative assessment – Module 3

5. Factorise the following polynomials: 5.2) 2 x 3 − x 2 − 15 x + 18 Continued ∴ 2 x3 − x 2 − 15 x + 18 2 = ( x − 2)(2 x + 3 x − 9) [Factorise] =( x − 2)(2 x − 3)( x + 3) 2x 3 −3x x 3 +6x 3x Downloaded from hlayiso.com Summative assessment – Module 3

6. If p + p − 2 is a factor of 2 f ( p ) = p − mp − 5 p + 8 p − n 4 3 2 Calculate the values of m and n Factorise p 2 + p − 2 to find factors ( p + p − 2 ) =0 2 ( p − 1)( p + 2 ) = 0 p − 1 =0 or p+2=0 p1 = 1 p 2 = −2 Downloaded from hlayiso.com Summative assessment – Module 3

6. Continued p − mp − 5 p + 8 p − n =f ( p ) 4 3 2 p1 1 = ⇒ f (1) 0 = [( p − 1) is a factor] ∴ (1) − m (1) − 5 (1) + 8 (1) − n = 4 3 2 0 −m − n + 4 =0 ...(1) p2 =−2 ⇒ f (−2) =0 [( p + 2 ) is a factor] ( −2 ) − m ( −2 ) − 5 ( −2 ) + 8 ( −2 ) − n = 4 3 2 0 8m − n − 20 = Downloaded from hlayiso.com 0 ...(2) Summative assessment – Module 3

6. Continued (1) − (2) : −m − n + 4 =0 ...(1) 8m − n − 20 =0 ...(2) −9m + 24 = 0 24 8 m= = 9 3 Substitute into (1) : −m − n + 4 =0 8 −  − n + 4 =0 3 4 n= 3 Downloaded from hlayiso.com Summative assessment – Module 3

7. If f ( x ) = x3 − 2 x 2 + ax − b is divided by ( x + 1), the remainder is 4, and when it is divided by ( x − 2 ) , the remainder is – 2. Calculate the values of a and b. x +1 =0 ⇒ x =−1 f ( −1) = 4 [Remainder is 4] x 3 − 2 x 2 + ax − b =f ( x ) ( −1) − 2 ( −1) + a ( −1) − b =4 3 2 −a − b − 1 − 2 − 4 =0 −a − b − 7 =0 ...(1) Downloaded from hlayiso.com Summative assessment – Module 3

7. Continued x−2=0 ⇒x=2 f ( 2 ) = −2 [Remainder is -2] x − 2 x + ax − b =f ( x ) 3 2 ( 2 ) − 2 ( 2 ) + a ( 2 ) − b =−2 3 2 8 − 8 + 2a − b + 2 =0 2a − b + 2 =0 ...(2) Downloaded from hlayiso.com Summative assessment – Module 3

7. Continued (1) − (2) : −a − b − 7 =0 ...(1) 2a − b + 2 =0 ...(2) −3a − 9 =0 9 a= = −3 −3 Substitute into (1) : −a − b − 7 =0 −(−3) − b − 7 =0 b = −4 Downloaded from hlayiso.com Summative assessment – Module 3

Mathematics NQF Level 2 Downloaded from hlayiso.com Mathematics NQF Level 4