{"id":15756,"date":"2022-05-23T03:30:43","date_gmt":"2022-05-22T22:00:43","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=15756"},"modified":"2023-11-10T10:30:28","modified_gmt":"2023-11-10T05:00:28","slug":"selina-icse-solutions-class-10-maths-remainder-factor-theorems","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/selina-icse-solutions-class-10-maths-remainder-factor-theorems\/","title":{"rendered":"Selina Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems"},"content":{"rendered":"
Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 8\u00a0Remainder and Factor Theorems<\/strong><\/p>\n Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> Question 13.<\/strong><\/span> Question 14.<\/strong><\/span> Question 15.<\/strong><\/span> Question 16.<\/strong><\/span> Question 17.<\/strong><\/span> Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 1.<\/strong><\/span> Question 2.<\/strong><\/span>Remainder and Factor Theorems Exercise 8A – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n
\n
\nSolution:<\/strong><\/span>
\nBy remainder theorem we know that when a polynomial f (x) is divided by x – a, then the remainder is f(a).
\n<\/p>\n
\n
\nSolution:<\/strong><\/span>
\n(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.
\n<\/p>\n
\nUse the Remainder Theorem to find which of the following is a factor of 2x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 5x – 6.
\n(i) x + 1
\n(ii) 2x – 1
\n(iii) x + 2
\nSolution:<\/strong><\/span>
\nBy remainder theorem we know that when a polynomial f (x) is divided by x – a, then the remainder is f(a).
\nLet f(x) = 2x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 5x – 6
\n(i) f (-1) = 2(-1)3<\/sup>\u00a0+ 3(-1)2<\/sup>\u00a0– 5(-1) – 6 = -2 + 3 + 5 – 6 = 0
\nThus, (x + 1) is a factor of the polynomial f(x).
\n
\nThus, (2x – 1) is not a factor of the polynomial f(x).
\n(iii) f (-2) = 2(-2)3<\/sup>\u00a0+ 3(-2)2<\/sup>\u00a0– 5(-2) – 6 = -16 + 12 + 10 – 6 = 0
\nThus, (x + 2) is a factor of the polynomial f(x).<\/p>\n
\n(i) If 2x + 1 is a factor of 2x2<\/sup>\u00a0+ ax – 3, find the value of a.
\n(ii) Find the value of k, if 3x – 4 is a factor of expression 3x2<\/sup>\u00a0+ 2x – k.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nFind the values of constants a and b when x – 2 and x + 3 both are the factors of expression x3<\/sup>\u00a0+ ax2<\/sup>\u00a0+ bx – 12.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nfind the value of k, if 2x + 1 is a factor of (3k + 2)x3<\/sup>\u00a0+ (k – 1).
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nFind the value of a, if x – 2 is a factor of 2x5<\/sup>\u00a0– 6x4<\/sup>\u00a0– 2ax3<\/sup>\u00a0+ 6ax2<\/sup>\u00a0+ 4ax + 8.
\nSolution:<\/strong><\/span>
\nf(x) = 2x5<\/sup>\u00a0– 6x4<\/sup>\u00a0– 2ax3<\/sup>\u00a0+ 6ax2<\/sup>\u00a0+ 4ax + 8
\nx – 2 = 0 \u21d2\u00a0 x = 2
\nSince, x – 2 is a factor of f(x), remainder = 0.
\n2(2)5<\/sup>\u00a0– 6(2)4<\/sup>\u00a0– 2a(2)3<\/sup>\u00a0+ 6a(2)2<\/sup>\u00a0+ 4a(2) + 8 = 0
\n64 – 96 – 16a + 24a + 8a + 8 = 0
\n-24 + 16a = 0
\n16a = 24
\na = 1.5<\/p>\n
\nFind the values of m and n so that x – 1 and x + 2 both are factors of x3<\/sup>\u00a0+ (3m + 1) x2<\/sup>\u00a0+ nx – 18.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nWhen x3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– kx + 4 is divided by x – 2, the remainder is k. Find the value of constant k.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nFind the value of a, if the division of ax3<\/sup>\u00a0+ 9x2<\/sup>\u00a0+ 4x – 10 by x + 3 leaves a remainder 5.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nIf x3<\/sup>\u00a0+ ax2<\/sup>\u00a0+ bx + 6 has x – 2 as a factor and leaves a remainder 3 when divided by x – 3, find the values of a and b.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nThe expression 2x3<\/sup>\u00a0+ ax2<\/sup>\u00a0+ bx – 2 leaves remainder 7 and 0 when divided by 2x – 3 and x + 2 respectively. Calculate the values of a and b.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nWhat number should be added to 3x3<\/sup>\u00a0– 5x2<\/sup>\u00a0+ 6x so that when resulting polynomial is divided by x – 3, the remainder is 8?
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nWhat number should be subtracted from x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 8x + 14 so that on dividing it with x – 2, the remainder is 10.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nThe polynomials 2x3<\/sup>\u00a0– 7x2<\/sup>\u00a0+ ax – 6 and x3<\/sup>\u00a0– 8x2<\/sup>\u00a0+ (2a + 1)x – 16 leaves the same remainder when divided by x – 2. Find the value of ‘a’.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nIf (x – 2) is a factor of the expression 2x3<\/sup>\u00a0+ ax2<\/sup>\u00a0+ bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nFind ‘a<\/span>‘ if the two polynomials ax3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 9 and 2x3<\/sup>\u00a0+ 4x + a, leave the same remainder when divided by x + 3.
\nSolution:<\/strong><\/span>
\n<\/p>\nRemainder and Factor Theorems Exercise 8B – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n
\nUsing the Factor Theorem, show that:
\n(i) (x – 2) is a factor of x3<\/sup>\u00a0– 2x2<\/sup>\u00a0– 9x + 18. Hence, factorise the expression x3<\/sup>\u00a0– 2x2<\/sup>\u00a0– 9x + 18 completely.
\n(ii) (x + 5) is a factor of 2x3<\/sup>\u00a0+ 5x2<\/sup>\u00a0– 28x – 15. Hence, factorise the expression 2x3<\/sup>\u00a0+ 5x2<\/sup>\u00a0– 28x – 15 completely.
\n(iii) (3x + 2) is a factor of 3x3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– 3x – 2. Hence, factorise the expression 3x3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– 3x – 2 completely.
\nSolution:<\/strong><\/span>
\n
\n<\/p>\n
\nUsing the Remainder Theorem, factorise each of the following completely.
\n(i)\u00a03x3\u00a0<\/sup>+ 2x2<\/sup>\u00a0\u2212 19x + 6
\n(ii) 2x3<\/sup>\u00a0+ x2<\/sup>\u00a0– 13x + 6
\n(iii) 3x3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– 23x – 30
\n(iv) 4x3<\/sup>\u00a0+ 7x2<\/sup>\u00a0– 36x – 63
\n(v) x3<\/sup>\u00a0+ x2<\/sup>\u00a0– 4x – 4
\nSolution:<\/strong><\/span>
\n
\n
\n<\/p>\n
\nUsing the Remainder Theorem, factorise the expression 3x3<\/sup>\u00a0+ 10x2<\/sup>\u00a0+ x – 6. Hence, solve the equation 3x3<\/sup>\u00a0+ 10x2<\/sup>\u00a0+ x – 6 = 0.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nFactorise the expression f (x) = 2x3<\/sup>\u00a0– 7x2<\/sup>\u00a0– 3x + 18. Hence, find all possible values of x for which f(x) = 0.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nGiven that x – 2 and x + 1 are factors of f(x) = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ ax + b; calculate the values of a and b. Hence, find all the factors of f(x).
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nThe expression 4x3<\/sup>\u00a0– bx2<\/sup>\u00a0+ x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.
\nSolution:<\/strong><\/span>
\n
\n<\/p>\n
\nIf x + a is a common factor of expressions f(x) = x2<\/sup>\u00a0+ px + q and g(x) = x2<\/sup>\u00a0+ mx + n;
\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nThe polynomials ax3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 3 and 2x3<\/sup>\u00a0– 5x + a, when divided by x – 4, leave the same remainder in each case. Find the value of a.
\nSolution:<\/strong><\/span>
\nLet f(x) = ax3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 3
\nWhen f(x) is divided by (x – 4), remainder = f(4)
\nf(4) = a(4)3<\/sup>\u00a0+ 3(4)2<\/sup>\u00a0– 3 = 64a + 45
\nLet g(x) = 2x3<\/sup>\u00a0– 5x + a
\nWhen g(x) is divided by (x – 4), remainder = g(4)
\ng(4) = 2(4)3<\/sup>\u00a0– 5(4) + a = a + 108
\nIt is given that f(4) = g(4)
\n64a + 45 = a + 108
\n63a = 63
\na = 1<\/p>\n
\nFind the value of ‘a’, if (x – a) is a factor of x3<\/sup>\u00a0– ax2<\/sup>\u00a0+ x + 2.
\nSolution:<\/strong><\/span>
\nLet f(x) = x3<\/sup>\u00a0– ax2<\/sup>\u00a0+ x + 2
\nIt is given that (x – a) is a factor of f(x).
\nRemainder = f(a) = 0
\na3<\/sup>\u00a0– a3<\/sup>\u00a0+ a + 2 = 0
\na + 2 = 0
\na = -2<\/p>\n
\nFind the number that must be subtracted from the polynomial 3y3<\/sup>\u00a0+ y2<\/sup>\u00a0– 22y + 15, so that the resulting polynomial is completely divisible by y + 3.
\nSolution:<\/strong><\/span>
\nLet the number to be subtracted from the given polynomial be k.
\nLet f(y) = 3y3<\/sup>\u00a0+ y2<\/sup>\u00a0– 22y + 15 – k
\nIt is given that f(y) is divisible by (y + 3).
\nRemainder = f(-3) = 0
\n3(-3)3<\/sup>\u00a0+ (-3)2<\/sup>\u00a0– 22(-3) + 15 – k = 0
\n-81 + 9 + 66 + 15 – k = 0
\n9 – k = 0
\nk = 9<\/p>\nRemainder and Factor Theorems Exercise 8C – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n
\nShow that (x – 1) is a factor of x3<\/sup>\u00a0– 7x2<\/sup>\u00a0+ 14x – 8. Hence, completely factorise the given expression.
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nUsing Remainder Theorem,\u00a0factorise:
\nx3<\/sup>\u00a0+ 10x2<\/sup>\u00a0– 37x + 26 completely.
\nSolution:<\/strong><\/span>