{"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 Concise Mathematics Class 10 ICSE Solutions Remainder and Factor Theorems<\/h2>\n

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 8\u00a0Remainder and Factor Theorems<\/strong><\/p>\n

Remainder and Factor Theorems Exercise 8A – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Question 1.<\/strong><\/span>
\n\"Selina
\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\"Selina<\/p>\n

Question 2.<\/strong><\/span>
\n\"Selina
\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\"Selina<\/p>\n

Question 3.<\/strong><\/span>
\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\"Selina
\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

Question 4.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 5.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 6.<\/strong><\/span>
\nfind the value of k, if 2x + 1 is a factor of (3k + 2)x3<\/sup>\u00a0+ (k – 1).
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 7.<\/strong><\/span>
\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

Question 8.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 9.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 10.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 11.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 12.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 13.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 14.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 15.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 16.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 17.<\/strong><\/span>
\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\"Selina<\/p>\n

Remainder and Factor Theorems Exercise 8B – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Question 1.<\/strong><\/span>
\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\"Selina
\n\"Selina<\/p>\n

Question 2.<\/strong><\/span>
\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\"Selina
\n\"Selina
\n\"Selina<\/p>\n

Question 3.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 4.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 5.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 6.<\/strong><\/span>
\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\"Selina
\n\"Selina<\/p>\n

Question 7.<\/strong><\/span>
\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\"Selina
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 8.<\/strong><\/span>
\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

Question 9.<\/strong><\/span>
\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

Question 10.<\/strong><\/span>
\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>\n

Remainder and Factor Theorems Exercise 8C – Selina Concise Mathematics Class 10 ICSE Solutions<\/h3>\n

Question 1.<\/strong><\/span>
\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\"Selina<\/p>\n

Question 2.<\/strong><\/span>
\nUsing Remainder Theorem,\u00a0factorise:
\nx3<\/sup>\u00a0+ 10x2<\/sup>\u00a0– 37x + 26 completely.
\nSolution:<\/strong><\/span>
\n\"\"<\/p>\n

Question 3.<\/strong><\/span>
\nWhen x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– mx + 4 is divided by x – 2, the remainder is m + 3. Find the value of m.
\nSolution:<\/strong><\/span>
\nLet f(x) = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– mx + 4
\nAccording to the given information,
\nf(2) = m + 3
\n(2)3<\/sup>\u00a0+ 3(2)2<\/sup>\u00a0– m(2) + 4 = m + 3
\n8 + 12 – 2m + 4 = m + 3
\n24 – 3 = m + 2m
\n3m = 21
\nm = 7<\/p>\n

Question 4.<\/strong><\/span>
\nWhat should be subtracted from 3x3<\/sup>\u00a0– 8x2<\/sup>\u00a0+ 4x – 3, so that the resulting expression has x + 2 as a factor?
\nSolution:<\/strong><\/span>
\nLet the required number be k.
\nLet f(x) = 3x3<\/sup>\u00a0– 8x2<\/sup>\u00a0+ 4x – 3 – k
\nAccording to the given information,
\nf (-2) = 0
\n3(-2)3<\/sup>\u00a0– 8(-2)2<\/sup>\u00a0+ 4(-2) – 3 – k = 0
\n-24 – 32 – 8 – 3 – k = 0
\n-67 – k = 0
\nk = -67
\nThus, the required number is -67.<\/p>\n

Question 5.<\/strong><\/span>
\nIf (x + 1) and (x – 2) are factors of x3<\/sup>\u00a0+ (a + 1)x2<\/sup>\u00a0– (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.
\nSolution:<\/strong><\/span>
\nLet f(x) = x3<\/sup>\u00a0+ (a + 1)x2<\/sup>\u00a0– (b – 2)x – 6
\nSince, (x + 1) is a factor of f(x).
\nRemainder = f(-1) = 0
\n(-1)3<\/sup>\u00a0+ (a + 1)(-1)2<\/sup>\u00a0– (b – 2) (-1) – 6 = 0
\n-1 + (a + 1) + (b – 2) – 6 = 0
\na + b – 8 = 0 …(i)
\nSince, (x – 2) is a factor of f(x).
\nRemainder = f(2) = 0
\n(2)3<\/sup>\u00a0+ (a + 1) (2)2<\/sup>\u00a0– (b – 2) (2) – 6 = 0
\n8 + 4a + 4 – 2b + 4 – 6 = 0
\n4a – 2b + 10 = 0
\n2a – b + 5 = 0 …(ii)
\nAdding (i) and (ii), we get,
\n3a – 3 = 0
\na = 1
\nSubstituting the value of a in (i), we get,
\n1 + b – 8 = 0
\nb = 7
\nf(x) = x3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– 5x – 6
\nNow, (x + 1) and (x – 2) are factors of f(x). Hence, (x + 1) (x – 2) = x2<\/sup>\u00a0– x – 2 is a factor of f(x).
\n\"Selina
\nf(x) = x3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– 5x – 6 = (x + 1) (x – 2) (x + 3)<\/p>\n

Question 6.<\/strong><\/span>
\nIf x – 2 is a factor of x2<\/sup>\u00a0+ ax + b and a + b = 1, find the values of a and b.
\nSolution:<\/strong><\/span>
\nLet f(x) = x2<\/sup>\u00a0+ ax + b
\nSince, (x – 2) is a factor of f(x).
\nRemainder = f(2) = 0
\n(2)2<\/sup>\u00a0+ a(2) + b = 0
\n4 + 2a + b = 0
\n2a + b = -4 …(i)
\nIt is given that:
\na + b = 1 …(ii)
\nSubtracting (ii) from (i), we get,
\na = -5
\nSubstituting the value of a in (ii), we get,
\nb = 1 – (-5) = 6<\/p>\n

Question 7.<\/strong><\/span>
\nFactorise x3<\/sup>\u00a0+ 6x2<\/sup>\u00a0+ 11x + 6 completely using factor theorem.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 8.<\/strong><\/span>
\nFind the value of ‘m’, if mx3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– 3 and x2<\/sup>\u00a0– mx + 4 leave the same remainder when each is divided by x – 2.
\nSolution:<\/strong><\/span>
\nLet f(x) = mx3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– 3
\ng(x) = x2<\/sup>\u00a0– mx + 4
\nIt is given that f(x) and g(x) leave the same remainder when divided by (x – 2). Therefore, we have:
\nf (2) = g (2)
\nm(2)3<\/sup>\u00a0+ 2(2)2<\/sup>\u00a0– 3 = (2)2<\/sup>\u00a0– m(2) + 4
\n8m + 8 – 3 = 4 – 2m + 4
\n10m = 3
\nm = 3\/10<\/p>\n

Question 9.<\/strong><\/span>
\nThe polynomial px3<\/sup>\u00a0+ 4x2<\/sup>\u00a0– 3x + q is completely divisible by x2<\/sup>\u00a0– 1; find the values of p and q. Also, for these values of p and q factorize the given polynomial completely.
\nSolution:<\/strong><\/span>
\nLet f(x) = px3<\/sup>\u00a0+ 4x2<\/sup>\u00a0– 3x + q
\nIt is given that f(x) is completely divisible by (x2<\/sup>\u00a0– 1) = (x + 1)(x – 1).
\nTherefore, f(1) = 0 and f(-1) = 0
\nf(1) = p(1)3<\/sup>\u00a0+ 4(1)2<\/sup>\u00a0– 3(1) + q = 0
\np + q + 1 = 0 …(i)
\nf(-1) = p(-1)3<\/sup>\u00a0+ 4(-1)2<\/sup>\u00a0– 3(-1) + q = 0
\n-p + q + 7 = 0 …(ii)
\nAdding (i) and (ii), we get,
\n2q + 8 = 0
\nq = -4
\nSubstituting the value of q in (i), we get,
\np = -q – 1 = 4 – 1 = 3
\nf(x) = 3x3<\/sup>\u00a0+ 4x2<\/sup>\u00a0– 3x – 4
\nGiven that f(x) is completely divisible by (x2<\/sup>\u00a0– 1).
\n\"Selina<\/p>\n

Question 10.<\/strong><\/span>
\nFind the number which should be added to x2<\/sup>\u00a0+ x + 3 so that the resulting polynomial is completely divisible by (x + 3).
\nSolution:<\/strong><\/span>
\nLet the required number be k.
\nLet f(x) = x2<\/sup>\u00a0+ x + 3 + k
\nIt is given that f(x) is divisible by (x + 3).
\nRemainder = 0
\nf (-3) = 0
\n(-3)2<\/sup>\u00a0+ (-3) + 3 + k = 0
\n9 – 3 + 3 + k = 0
\n9 + k = 0
\nk = -9
\nThus, the required number is -9.<\/p>\n

Question 11.<\/strong><\/span>
\nWhen the polynomial x3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– 5ax – 7 is divided by (x – 1), the remainder is A and when the polynomial x3<\/sup>\u00a0+ ax2<\/sup>\u00a0– 12x + 16 is divided by (x + 2), the remainder is B. Find the value of ‘a’ if 2A + B = 0.
\nSolution:<\/strong><\/span>
\nIt is given that when the polynomial x3<\/sup>\u00a0+ 2x2<\/sup>\u00a0– 5ax – 7 is divided by (x – 1), the remainder is A.
\n(1)3<\/sup>\u00a0+ 2(1)2<\/sup>\u00a0– 5a(1) – 7 = A
\n1 + 2 – 5a – 7 = A
\n– 5a – 4 = A …(i)
\nIt is also given that when the polynomial x3<\/sup>\u00a0+ ax2<\/sup>\u00a0– 12x + 16 is divided by (x + 2), the remainder is B.
\nx3<\/sup>\u00a0+ ax2<\/sup>\u00a0– 12x + 16 = B
\n(-2)3<\/sup>\u00a0+ a(-2)2<\/sup>\u00a0– 12(-2) + 16 = B
\n-8 + 4a + 24 + 16 = B
\n4a + 32 = B …(ii)
\nIt is also given that 2A + B = 0
\nUsing (i) and (ii), we get,
\n2(-5a – 4) + 4a + 32 = 0
\n-10a – 8 + 4a + 32 = 0
\n-6a + 24 = 0
\n6a = 24
\na = 4<\/p>\n

Question 12.<\/strong><\/span>
\n(3x + 5) is a factor of the polynomial (a – 1)x3<\/sup>\u00a0+ (a + 1)x2<\/sup>\u00a0– (2a + 1)x – 15. Find the value of ‘a’, factorise the given polynomial completely.
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 13.<\/strong><\/span>
\nWhen divided by x – 3 the polynomials x3<\/sup>\u00a0– px2<\/sup>\u00a0+ x + 6 and 2x3<\/sup>\u00a0– x2<\/sup>\u00a0– (p + 3) x – 6 leave the same remainder. Find the value of ‘p’.
\nSolution:<\/strong><\/span>
\nIf (x – 3) divides f(x) = x3<\/sup>\u00a0– px2<\/sup>\u00a0+ x + 6, then,
\nRemainder = f(3) = 33<\/sup>\u00a0– p(3)2<\/sup>\u00a0+ 3 + 6 = 36 – 9p
\nIf (x – 3) divides g(x) = 2x3<\/sup>\u00a0– x2<\/sup>\u00a0– (p + 3) x – 6, then
\nRemainder = g(3) = 2(3)3<\/sup>\u00a0– (3)2<\/sup>\u00a0– (p + 3) (3) – 6 = 30 – 3p
\nNow, f(3) = g(3)
\n\u21d2 36 – 9p = 30 – 3p
\n\u21d2 -6p = -6
\n\u21d2 \u00a0p = 1<\/p>\n

Question 14.<\/strong><\/span>
\nUse the Remainder Theorem to factorise the following expression:
\n2x3<\/sup>\u00a0+ x2<\/sup>\u00a0– 13x + 6
\nSolution:<\/strong><\/span>
\n\"Selina<\/p>\n

Question 15.
\n<\/strong><\/span>Using remainder theorem, find the value of k if on dividing 2x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0–\u00a0kx\u00a0+ 5 by x – 2, leaves a remainder 7.
\nSolution:<\/strong><\/span>
\nLet f(x) =\u00a02x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0–\u00a0kx\u00a0+ 5
\nUsing Remainder Theorem, we have
\nf(2) = 7
\n\u2234\u00a02(2)3<\/sup>\u00a0+ 3(2)2<\/sup>\u00a0– k(2) + 5 = 7
\n\u2234\u00a016 + 12 – 2k + 5 = 7
\n\u2234\u00a033 – 2k = 7
\n\u2234\u00a02k = 26
\n\u2234\u00a0k = 13<\/p>\n

Question 16.
\n<\/strong><\/span>What must be subtracted from 16x3<\/sup>\u00a0– 8x2<\/sup>\u00a0+ 4x + 7 so that the resulting expression has 2x + 1 as a factor?
\nSolution:<\/span>
\n<\/strong><\/span>Here, f(x) =\u00a016x3<\/sup>\u00a0– 8x2<\/sup>\u00a0+ 4x + 7
\nLet the number subtracted be k from the given polynomial f(x).
\nGiven that 2x + 1 is a factor of f(x).
\n\"Selina
\nTherefore 1\u00a0must be subtracted from 16x3<\/sup>\u00a0– 8x2<\/sup>\u00a0+ 4x + 7 so that the resulting expression has 2x + 1 as a factor.<\/p>\n

More Resources for Selina Concise Class 10 ICSE Solutions<\/strong><\/p>\n