Find values to fill these boxes and complete the identity: ◻ x 2 − ◻ x − ◻ = ◻ x 2 + ◻ x − 4 x − 14 = 5 x ( 2 x + 7 ) − ◻ ( 2 x + 7 ) = ( ◻ x − ◻ ) ( ◻ x + ◻ )

Given, □ x 2 - □ x - □ = □ x 2 + □ x - 4 x - 14 = 5 x ( 2 x + 7 ) - □ ( 2 x + 7 ) = ( □ x - □ ) ( □ x + □ ) Let us name the empty boxes with variables, a , b , c , d , e , f , g , h , i , and j as follows: a x 2 - b x - c = d x 2 + e x - 4 x - 14 . . . . i = 5 x 2 x + 7 - f 2 x + 7 . . . . i i = g x - h i x + j . . . . i i i Considering L.HS. a x 2 - b x - c = d x 2 + e x - 4 x - 14 = d x 2 + e x - 2 2 x + 7 . . . . . i v = 5 x ( 2 x + 7 ) - 2 ( 2 x + 7 ) (given) . . . . v = 5 x - 2 2 x + 7 . . . . . v i Now, comparing equations i i , and v we have f = 2 , g = 5 , h = 2 , i = 2 , j = 7 From R.H.S. of equation v we have 5 x 2 x + 7 - 2 2 x + 7 = 10 x 2 + 35 x - 4 x - 14 . . . . v i i From equations i v , and v i i we have d = 10 , and e = 35 Since a x 2 - b x - c = d x 2 + e x - 4 x - 14 = 10 x 2 + 35 x - 4 x - 14 = 10 x 2 + 31 x - 14 ⇒ a = 10 , b = - 31 , and c = 14 Hence, the values in the empty boxes are as follows: 10 x 2 − - 31 x − 14 = 10 x 2 + 35 x − 4 x − 14 = 5 x ( 2 x + 7 ) − 2 ( 2 x + 7 ) = ( 5 x − 2 ) ( 2 x + 7 )

Find values to fill these boxes and complete the identity: 4 x 2 + 27 x + ◻ = ◻ x 2 + 3 x + ◻ x + ◻ = x ( 4 x + 3 ) + ◻ ( 4 x + 3 ) = ( x + ◻ ) ( ◻ x + ◻ )

The given steps are clearly that of middle term factorisation of an algebraic expression. From the third step of the given factorisation, it is evident that one of the factors would be ( 4 x + 3 ) From the second step as given in question, it is evident that the middle term is to be expressed as the sum ( 3 x + 24 x ) . This is because, the second term is 27 x , as evident from the first step. Thus, the other factor turns out to be ( x + 6 ) , since we can take ( 4 x + 3 ) common from the third step and 6 is the multiple of ( 4 x + 3 ) in the second term of this step. So, the factorised form of the expression is = ( x + 6 ) ( 4 x + 3 ) We can clearly see that the third term in the first step of the question is 18 . So, the third term of first step is 18 . The first, second and third entities respectively in the blanks of the second step are 4 , 24 a n d 18 . The entity in the blank space of third step is 6 . The first, second and third entities respectively in the blank spaces of the fourth step are 6 , 4 a n d 3 .

E M B I B E

MYP Mathematics A concept based approach 4&5 Standard>Patterns>Chapter 17 - Back to the Beginning (Using Patterns to Work Backwards)>Practice 5

Rose Harrison and Clara Huizink Solutions for Exercise 17: Practice 5

Author:Rose Harrison & Clara Huizink

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison and Clara Huizink Solutions for Exercise 17: Practice 5

Attempt the practice questions from Exercise 17: Practice 5 with hints and solutions to strengthen your understanding. MYP Mathematics A concept based approach 4&5 Standard solutions are prepared by Experienced Embibe Experts.