Remainder and factor theorem class 10 icse

Maths Formulas for Class 10 is one of the best materials that students can get, as it will help them to concentrate better and reduce the level of stress that students face during the furious year. Maths formulas are essential for conceptual understanding and scoring good marks, and for Revision, a list of maths formulas are always considered the best when your exams are coming. We give ICSE Maths Formulas List for Class 10 in a straightforward, free downloadable PDF design for the students to figure out a better understanding of the solutions. On this page, we have provided Selina Maths Class 10 formulas for all chapters which are in 1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: Maths Formulas for ICSE Class 10 Goods And Service Tax Some Terms Related To Gst: GST – Goods and Service Tax Before GST, there were 13 different indirect taxes in India. These 13 indirect , taxes are replaced by GST (Goods and Service Tax). GST is levised by the Central and State Government and provide for a single and streamlined process. GST is a common tax for goods and service. The tax will be implemented at every step of value creation. CGST: The Central GST or CGST is tax collected by the Central, when the sales are within the. SGST: The State GST or SGST is tax collected by theState, when the sales are within states. the states. IGST: The Integrated GST or IGST is tax collected by the Central, when the sales are outside the states. Rates of gst: The GST is levied on multiple rates fro...

Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems Exercise 8(B) Using the factor theorem to factorise the given polynomial is the main concept discussed in this exercise. The students wishing to strengthen their problem-solving abilities on this chapter and others can use the Access other exercises of Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems Access Selina Solutions Concise Maths Class 10 Chapter 8 Remainder and Factor Theorems Exercise 8(B) 1. Using the Factor Theorem, show that: (i) (x – 2) is a factor of x 3 – 2x 2 – 9x + 18. Hence, factorise the expression x 3 – 2x 2 – 9x + 18 completely. (ii) (x + 5) is a factor of 2x 3 + 5x 2 – 28x – 15. Hence, factorise the expression 2x 3 + 5x 2 – 28x – 15 completely. (iii) (3x + 2) is a factor of 3x 3 + 2x 2 – 3x – 2. Hence, factorise the expression 3x 3 + 2x 2 – 3x – 2 completely. Solution: (i) Here, f(x) = x 3 – 2x 2 – 9x + 18 So, x – 2 = 0 ⇒ x = 2 Thus, remainder = f(2) = (2) 3 – 2(2) 2 – 9(2) + 18 = 8 – 8 – 18 + 18 = 0 Therefore, (x – 2) is a factor of f(x). Now, performing division of polynomial f(x) by (x – 2) we have Thus, x 3 – 2x 2 – 9x + 18 = (x – 2) (x 2 – 9) = (x – 2) (x + 3) (x – 3) (ii) Here, f(x) = 2x 3 + 5x 2 – 28x – 15 So, x + 5 = 0 ⇒ x = -5 Thus, remainder = f(-5) = 2(-5) 3 + 5(-5) 2 – 28(-5) – 15 = -250 + 125 + 140 – 15 = -265 + 265 = 0 Therefore, (x + 5) is a factor of f(x). Now, performing division of polynomial ...

Exercise 8(A) 1. Find, in each case, the remainder when: (i) x 4– 3x 2+ 2x + 1 is divided by x – 1. (ii) x 3+ 3x 2– 12x + 4 is divided by x – 2. (ii) x 4+ 1 is divided by x + 1. Solution From remainder theorem, we know that when a polynomial f (x) is divided by (x – a), then the remainder is f(a). (i) Given, f(x) = x 4– 3x 2+ 2x + 1 is divided by x – 1 So, remainder = f(1) = (1) 4– 3(1) 2+ 2(1) + 1 = 1 – 3 + 2 + 1 = 1 (ii) Given, f(x) = x 3+ 3x 2– 12x + 4 is divided by x – 2 So, remainder = f(2) = (2) 3+ 3(2) 2– 12(2) + 4 = 8 + 12 – 24 + 4 = 0 (iii) Given, f(x) = x 4+ 1 is divided by x + 1 So, remainder = f(-1) = (-1) 4+ 1 = 2 2. Show that: (i) x – 2 is a factor of 5x 2+ 15x – 50 (ii) 3x + 2 is a factor of 3x 2– x – 2 Solution (x – a) is a factor of a polynomial f(x) if the remainder, when f(x) is divided by (x – a), is 0, i.e., if f(a) = 0. (i) f(x) = 5x 2+ 15x – 50 f(2) = 5(2) 2+ 15(2) – 50 = 20 + 30 – 50 = 0 As the remainder is zero for x = 2 Thus, we can conclude that (x – 2) is a factor of 5x 2+ 15x – 50 (ii) f(x) = 3x 2– x – 2 f(-2/3) = 3(-2/3) 2– (-2/3) – 2 = 4/3 + 2/3 – 2 = 2 – 2 = 0 As the remainder is zero for x = -2/3 Thus, we can conclude that (3x + 2) is a factor of 3x 2– x – 2 3. Use the Remainder Theorem to find which of the following is a factor of 2x 3+ 3x 2– 5x – 6. (i) x + 1 (ii) 2x – 1 (iii) x + 2 Solution From remainder theorem we know that when a polynomial f (x) is divided by x – a, then the remainder is f(a). Here, f(x) = 2x 3+ 3x 2– 5x – 6 (i) f (-...

Question 1: Show is a factor of . Factorize the polynomial. Answer: For , Remainder: Hence is a factor of • • • • • • • Hence Question 2: Using remainder theorem, factorize completely. [2014] Answer: For , Remainder: Hence is a factor of • • • • • • • Hence Question 3: Find the value of when is divided by , the remainder is . Answer: When Remainder Question 4: What should be subtracted from , so that the resulting expression has as a factor. Answer: Let be subtracted When Remainder Question 5: If and are factors of , find the values of . Factorize the polynomial also. Answer: When Remainder: ……………… i) When Remainder: ……………… ii) Solving i) and ii) we get Substituting the values in the polynomial we get Given is a factor of • • • • • • • Hence Question 6: If is a factor of and , find the values of . Answer: When Remainder: Given Therefore Solving for Hence Question 7: Using remainder theorem, factorize completely. Answer: For , Remainder: Hence is a factor of • • • • • • • Hence Question 8: Find the value of , if and leave the same remainder when each is divided by . Answer: When Remainder 1 Remainder 2 Given Remainder 1 = Remainder 2 Question 9: The polynomial is completely divisible by . Find the value of . Also for these values of , factorize the given polynomial completely. Answer: is a factor of is a factor of are factors of When …………… i) When …………… ii) Solving i) and ii) we get Question 10: Find the number which should be added to so that the resulting polynomials comp...