Scoring well in the ICSE Class 10 Maths exam requires students to have a good understanding of the concepts and formulas. One way to prepare for the exam is by practicing the previous year papers. By solving these papers, students can get familiarized with the exam pattern and the types of questions asked in the exam. This can help them to approach the exam with confidence and reduce exam pressure. At our website, students can easily access the ICSE Class 10 Maths Previous Year Questions with solutions from 2010 to 2023.
ICSE Class 10 Maths Previous Year Questions with Solutions Chapter Factorisation 2010 to 2023:
The past year questions from Factorisation of Polynomials available on our website are from the years 2023, 2022, 2020, 2019, 2018, 2017, 2016, 2015, 2014, 2013, 2012, 2011 and 2010. Each sum is solved in detail that helps students understand the marking pattern.
Moreover, practicing the ICSE Class 10 Previous Years Questions Mathematics Chapter-Factorisation can be particularly beneficial for students. Factorisation is an important topic in Maths and requires a good understanding of formulas and concepts. By practicing the previous year questions related to factorisation, students can get a good grasp of the topic and be better prepared for the exam.
ICSE Class 10 Maths Previous Years Questions Chapter-Factorisation
Q1. Factorize completely using factor theorem: 2x3 – x2 – 13x – 6 [2023]
Answer: (x+2)(x-3)(2x+1)
Step-by-step Explanation:
\[P(x)=\;2x^3\;–\;x^2\;-\;13x\;–\;6\;\;\\Let\;x=-2,\;\\the\;value\;of\;f(x)\;will\;be\;\\f(-2)=2{(-2)}^3\;-{(-2)}^2\;\;–13(-2)-6\\\;\;\;\;\;\;\;\;\;\;=-16-4+26-6\\\;\;\;\;\;\;\;\;\;\;=0\;\\As\;f(-2)=0,\;so\;(x+2)\;is\;a\;factor\;of\;f(x).\;\\Now,\;performing\;long\;division\;we\;have\;Thus,\;\\\]
\[P(x)=\;2x^3\;–\;x^2\;-\;13x\;–\;6\;\;\\Let\;x=-2,\;\\the\;value\;of\;f(x)\;will\;be\;\\f(-2)=2{(-2)}^3\;-{(-2)}^2\;\;–13(-2)-6\\\;\;\;\;\;\;\;\;\;\;=-16-4+26-6\\\;\;\;\;\;\;\;\;\;\;=0\;\\As\;f(-2)=0,\;so\;(x+2)\;is\;a\;factor\;of\;f(x).\;\\Now,\;performing\;long\;division\;we\;have\;Thus,\;\\f(x)=(x+2)(2x^2\;-5x–3)\;\\\;\;\;\;\;\;=(x+2)\lbrack2x^2-6x+x–3\rbrack\;\\\;\;\;\;\;\;=(x+2)\lbrack2x(x-3)+1(x-3)\rbrack\;\\\;\;\;\;\;\;=(x+2)\lbrack(2x+1)(x-3)\rbrack\;\\\;\;\;\;\;\;=(x+2)(2x+1)(x-3)\]
Q2. Find the value of ‘a’ if x -a is a factor of the polynomial 3x3 + x2 – ax – 81. [4] [2023]
Answer: a=3
Step-by-step Explanation:
\[x-a=0\\x=aand,\\p(x)=3x^3+x^2-ax-81\\substituting\;x=a\;in\;p(x)\;we\;get,\\3a^3+a^2-a^2-81=0\\3a^3-81=0\\3a^3=81\\a^3=27\\a=3\]
Q3. If x -2 is a factor of x3 – kx -12, then the value of k is:
(a) 3
(b) 2
(c) -2
(d) -3 [2023]
Answer: (c) -2
Step-by-step Explanation:
\[P(x)=\;x^3\;-\;kx\;-12\\(x-2)is\;a\;factor\;of\;P(x)\;\;\\So,\;2\;is\;the\;zero\;of\;the\;polynomial\;\;\\Substitute\;x\;=\;2\;in\;P(x)\\x^3-kx-12=0\\2^3-k.2-12=0\\8-2k-12=0\\-2k-4=0\\-2k=4\\k=-2\]
Q4. If (x + 2) is a factor of the polynomial x3 – kx2 – 5x + 6 then the value of k is: [1]
(a) 1
(b) 2
(c) 3
(d) -2 [2021 Semester-1]
Answer: (b) 2
Step-by-step Explanation:
\[P(x)=\;x^3\;-\;kx^2\;-\;5x\;+\;6\;\\(x+2)is\;a\;factor\;of\;P(x)\;\;\\So,\;-2\;is\;the\;zero\;of\;the\;polynomial\;\;\\Substitute\;x\;=\;-2\;in\;P(x)\\x^3-kx^2\;-\;5x\;+\;6=0\\{(-2)}^3-k.{(-2)}^2-5.(-2)+6=0\\-8-4k+10+6=0\\-4k+8=0\\-4k=-8\\k=2\]
Q5. The polynomial x3 – 2x2 + ax + 12 when divided by (x + 1) leaves a remainder 20, then ‘a’ is equal to: [1]
(a) – 31
(b) 9
(c) 11
(d) – 11 [2021 Semester-1]
Answer: (d) -11
Step-by-step Explanation:
\[x+1=0\\x=-1and,\\p(x)=x^3-2x^2+ax+12\\By\;remainder\;theorem,\\p(-1)=20\\substituting\;x=-1\;in\;p(x)\;we\;get,\\{(-1)}^3-2.{(-1)}^2+a.(-1)+12=20\\-1-2-a+12=20\\9-a=20\\-a=11\\a=-11\]
Q6. (x + 2) and (x + 3) are two factors of the polynomial x3 + 6x2 + 11x + 6. If this polynomial is completely factorised the result is: [2]
(a) (x – 2)(x + 3)(x + 1)
(b) (x + 2)(x – 3)(x – 1)
(c) (x + 2)(x + 3)(x – 1)
(d) (x + 2)(x + 3)(x + 1) [2021 Semester-1]
Answer: (d) (x+2)(x+3)(x+1)
Step-by-step Explanation:
\[(x+2)(x+3)\;=\;x^2+2x+3x+6\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\;x^2+5x+6\]
\begin{array}{r}
x+1\phantom{)} \\
x^2+5x+6{\overline{\smash{\big)}\,x^3+6x^2+11x+6\phantom{)}}}\\
\underline{-~\phantom{(}(x^3+5x^2+6x)\phantom{-b)}}\\
0+x^2+5x+6\phantom{)}\\
\underline{-~\phantom{()}(x^2+5x+6)}\\
0\phantom{)}
\end{array}
\[Therefore,\;p(x)=(x+2)(x+3)\;(x+1)\]
Q7. What must be added to the polynomial 2𝑥3 − 3𝑥2 − 8𝑥, so that it leaves a remainder 10 when divided by 2x + 1? [2020]
Answer: 7
Step-by-step Explanation:
\[Let\;a\;must\;be\;added\;to\;the\;polynomial.\\Therefore,\;p(x)=2x^3-3x^2-8x+a\\The\;polynomial\;is\;divided\;by\;(2x+1)\\So,\;2x+1=0\\x=-\frac12\\Therefore,\;by\;remainder\;theorem,\\p(-\frac12)=10\\2{(-\frac12)}^3-3{(-\frac12)}^2-8(-\frac12)+a=10\\2.(-\frac18)-3.\frac14+\frac82+a=10\\-\frac14-\frac34+4+a=10\\a=10-4+\frac14+\frac34\\a=6+\frac14+\frac34\\a=\frac{24+1+3}4\\a=\frac{28}4\\a=7\\\]
Q8. Use factor theorem to factorise 6𝑥3 + 17𝑥2 + 4𝑥 − 12 completely. [2020]
Answer: (x+2)(2x+3)(3x-2)
Step-by-step Explanation:
\[p(x)=6x^3+17x^2\;+4\operatorname{𝑥}-12\\Taking\;x=-2\;we\;have,\\p(-2)\;=6.{(-2)}^3+17.{(-2)}^2+4.(-2)-12\\\;\;\;\;\;\;\;\;\;\;\;\;=\;-48+68-8-12\\\;\;\;\;\;\;\;\;\;\;\;\;=-68+68\\\;\;\;\;\;\;\;\;\;\;\;=\;0\\Therefore,\;(x+2)\;is\;a\;factor\;of\;p(x).\\dividing\;p(x)\;by\;(x+2)\;we\;have,\\\]
\begin{array}{r}
6x^2+5x-6\phantom{)} \\
x+2{\overline{\smash{\big)}\,6x^3+17x^2+4x-12\phantom{)}}}\\
\underline{-~\phantom{(}(6x^3+12x^2)\phantom{-b)}}\\
0+5x^2+4x-12\phantom{)}\\
\underline{-~\phantom{()}(5x^2+10x)}\\
0-6x-12\phantom{)}\\
\underline{-~\phantom{()}(-6x-12)}\\
0\phantom{)}
\end{array}
\[6x^2+5\operatorname{𝑥}-6\\=6x^2+9\operatorname{𝑥}-4x-6\\=3x(2x+3)-2(2x+3)\\=(3x-2)(2x+3)\\Therefore\;p(x)=\;(x+2)(3x-2)(2x+3)\\\\\]
Q9. Using the factor theorem, show that (x – 2) is a factor of x3 + x2 – 4x – 4. [3]
Hence, factorise the polynomial completely. [2019]
Answer: (x-2)(x+2)(x+1)
Step-by-step Explanation:
\[f(x)=\;x^3\;+\;x^2\;–\;4x\;–\;4.\\Let\;x-2\;=\;0\\x=2\\Therefore,\;\\f(2)={(2)}^3+{(2)}^2-4.2-4\\\;\;\;\;\;\;=\;8+4-8-4\\\;\;\;\;\;\;=\;0\\Hence,\;x-2\;is\;a\;factor\;of\;f(x).\\Dividing\;f(x)\;by\;(x-2),\;we\;have,\\\\\\\\\]
\begin{array}{r}
x^2+3x+2\phantom{)} \\
x-2{\overline{\smash{\big)}\,x^3+x^2-4x-4\phantom{)}}}\\
\underline{-~\phantom{(}(x^3-2x^2)\phantom{-b)}}\\
0+3x^2-4x-4\phantom{)}\\
\underline{-~\phantom{()}(3x^2-6x)}\\
0+2x-4\phantom{)}\\
\underline{-~\phantom{()}(2x-4)}\\
0\phantom{)}
\end{array}
\[f(x)=\;(x-2)(x+2)(x+1)\\\\\\\\\]
Q10. Using the Remainder Theorem find the remainders obtained when x3 + (kx + 8)x + k is divided by x + 1 and x – 2. Hence, find k if the sum of the two remainders is 1. [3] [2019]
Answer: k=-2
Step-by-step Explanation:
\[f(x)=\;x^3\;+\;(kx\;+\;8)x\;+\;k\;\\g(x)=x+1\\So,\;x=-1\\u\sin g\;the\;remainder\;theorem,\\f(-1)=Remainder_1\\{(-1)}^3+\{k.(-1)+8\}.(-1)+k\\-1+k-8+k\\Remainder_1=2k-9\\Now,\;h(x)=x-2\\Therefore,\;x=2\\f(2)=Remainder_2\\{(2)}^3+(k.2+8).2+k\\8+4k+16+k\\Remainder_2=5k+24\\Given\;that,\\(2k-9)+(5k+24)=1\\7k+15=1\\7k=-14\\k=-2\\\\\\\\\]
Q11. If (x + 2) and (x + 3) are factors of x3 + ax + b, find the values of ‘a’ and ‘b’. [3] [2018]
Answer: a =-19 , b = -30.
Step-by-step Explanation:
\[f(x)=\;x^3+ax+b\\Given,\;(x+2)\;is\;a\;factor\;of\;f(x).\\By\;factor\;theorem,\\f(-2)=0\\{(-2)}^3+a.(-2)+b=0\\-8-2a+b=0\\-2a+b=8…….(1)\\Also\;given\;(x+3)\;is\;a\;factor\;of\;f(x)\\f(-3)=0\\{(-3)}^3+a.(-3)+b=0\\-27-3a+b=0\\-3a+b=27…….(2)\\Subtracting\;(1)\;from\;(2)\;we\;have,\\-a=19\\a=-19\\substituting\;a=-19\;in\;(1)\;we\;have\\-2\times(-19)+b=8\\38+b=8\\b=-30\\Hence,\;a=-19\;and\;b=-30.\\\\\\\\\\\]
Q12. Use Remainder theorem to factorize the following polynomial: [3]
2x3 + 3x2 – 9x – 10. [2018]
Answer: (x-2)(x+1)(2x+5)
Step-by-step Explanation:
\[f(x)=\;2x^3+3x^2-9x-10\\Taking\;x=2\;we\;have,\\2.{(2)}^3+3.{(2)}^2-9.(2)-10\\=16+12-18-10\\=28-28\\=0\\Therefore,\;(x-2)\;is\;a\;factor\;of\;f(x).\\Dividing\;f(x)\;by\;(x-2),\;we\;have,\\\\\\\\\\\\\]
\begin{array}{r}
2x^2+7x+5\phantom{)} \\
x-2{\overline{\smash{\big)}\,2x^3+3x^2-9x-10\phantom{)}}}\\
\underline{-~\phantom{(}(2x^3-4x^2)\phantom{-b)}}\\
0+7x^2-9x-10\phantom{)}\\
\underline{-~\phantom{()}(7x^2-14x)}\\
0+5x-10\phantom{)}\\
\underline{-~\phantom{()}(5x-10)}\\
0\phantom{)}
\end{array}
\[2x^2+7x+5\\=\;2x^2+5x+2x+5\\=\;x(2x+5)+1(2x+5)\\=\;(2x+5)(x+1)\\Hence,\;f(x)=(x-2)(x+1)(2x+5)\\\\\\\\\\\\\\\]
Q13. What must be subtracted from 16x3 – 8x2 + 4x + 7 so that the resulting expression has 2x + 1 as a factor? [3] [2017]
Answer: 1
Step-by-step Explanation:
\[Let\;a\;be\;subtracted.\;Therefore,\\f(x)=\;16x^3–8x^2+4x+7-a\\g(x)=2x+1\\So,\;x=-\frac12\\By\;factor\;theorem,\\f(-\frac12)=0\\16{(-\frac12)}^3–8{(-\frac12)}^2+4(-\frac12)+7-a=0\\\frac{-16}8-\frac84-\frac42+7-a=0\\-2-2-2+7-a=0\\1-a=0\\a=1\]
Q14. Using remainder theorem, find the value of k, if on dividing 2x3 + 3x2 – kx + 5 by x-2, leaves a remainder 7. [3] [2016]
Answer: k=13
Step-by-step Explanation:
\[Let\;a\;be\;subtracted.\;Therefore,\\f(x)=\;2x^3+3x^2-kx+5\\g(x)=x-2\\So,\;x=2\\By\;remainder\;theorem,\\f(2)=7\;\\2{(2)}^3+3{(2)}^2-k.2+5=7\\16+12-2k+5=7\\33-2k=7\\-2k=7-33\\-2k=-26\\k=13\]
Q15. Find ‘a’ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leaves the same remainder when divided by x + 3. [3] [2015]
Answer: a=3
Step-by-step Explanation:
\[The\;given\;polynomials\;are\\p(x)=ax^3\;+\;3x^2\;–\;9\;and\;\\q(x)=\;2x^3\;+\;4x\;+\;a\\Given\;that\;p(x)\;and\;q(x)\;leave\;the\;same\;remainder\;when\;divided\;by\;x+3.\\Thus\;by\;remainder\;theorem,\\p(-3)=q(-3)\\\Rightarrow a{(-3)}^3\;+\;3{(-3)}^2\;–\;9\;=\;2{(-3)}^3\;+\;4(-3)\;+\;a\\\Rightarrow-27a+27-9=-54-12+a\\\Rightarrow-27a-a=-54-12-27+9\\\Rightarrow-28a=-93+9\\\Rightarrow-28a=-84\\\Rightarrow a=3\\\]
Q16. Using the Remainder and Factor Theorem, factorise the following polynomial:
x3 + 10x2 – 37x + 26. [3] [2014]
Answer: (x-1)(x-2)(x+13)
Step-by-step Explanation:
\[\\f(x)=x^3+10x^2–37x+26.\;\\Let\;x=1\\f(1)={(1)}^3+10{(1)}^2–37(1)+26\\=\;1+10-37+26\\=\;0\\Therefore,\;By\;factor\;theorem,\\(x-1)\;is\;a\;factor\;of\;f(x).\\Dividing\;f(x)\;by\;x-1\;we\;have,\\\\\]
\begin{array}{r}
x^2+11x-26\phantom{)} \\
x-1{\overline{\smash{\big)}\,x^3+10x^2-37x+26\phantom{)}}}\\
\underline{-~\phantom{(}(x^3-x^2)\phantom{-b)}}\\
0+11x^2-37x+26\phantom{)}\\
\underline{-~\phantom{()}(11x^2-11x)}\\
0-26x+26\phantom{)}\\
\underline{-~\phantom{()}(-26x+26)}\\
0\phantom{)}
\end{array}
\[x^2+11x-26\\=\;x^2+13x-2x-26\\=\;x(x+13)-2(x+13)\\=\;(x+13)(x-2)\\Therefore,\;f(x)=(x-1)(x-2)(x+13)\]
Q17. If (x – 2) is a factor of the expression 2x3 + ax2 + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b. [3] [2013]
Answer: a = 5 ; b = -11
Step-by-step Explanation:
\[f(x)=2x^3\;+\;ax^2\;+\;bx\;–\;14\;\\Given,\;(x-2)\;is\;a\;factor\;of\;f(x).\\By\;factor\;theorem,\\f(2)=0\\2{(2)}^3\;+\;a{(2)}^2\;+\;b(2)\;–\;14\;=0\\16+4a+2b-14=0\\4a+2b=-2\\2a+b=-1……\;(1)\\Given,\;when\;f(x)\;is\;divided\;by\;(x-3),\;it\;leaves\;52\;as\;remainder.\\Therefore,\;By\;remainder\;theorem,\\f(3)=52\\2{(3)}^3\;+\;a{(3)}^2\;+\;b(3)\;–\;14\;=52\\54+9a+3b-14=52\\9a+3b=52+14-54\\3(3a+b)=12\\3a+b=4……\;(2)\\subtracting\;(1)\;by\;(2)\;we\;get,\\a=5\\Substituting\;a=5\;in\;(1)\\2a+b=-1\\2\times5+b=-1\\10+b=-1\\b=-11\\Hence,\;a\;=\;5\;and\;b\;=\;-11\]
Q18. Using the Remainder Theorem factorise completely the following polynomial:
3x3 + 2x2 – 19x + 6. [3] [2012]
Answer: (x-2)(x+3)(3x-1)
Step-by-step Explanation:
\[f(x)=3x^3\;+\;2x^2\;–\;19x\;+\;6.\\Taking\;x=2\;we\;have,\\f(2)=3{(2)}^3+2{(2)}^2-19\times2+6\\\;\;\;\;\;\;=24+8-38+6\\\;\;\;\;\;\;=\;38-38\\\;\;\;\;\;\;=\;0\\Therefore,\;(x-2)\;is\;a\;factor\;of\;f(x).\\Dividing\;f(x)\;by\;(x-2),\;we\;have,\\\;\]
\begin{array}{r}
3x^2+8x-3\phantom{)} \\
x-2{\overline{\smash{\big)}\,3x^3+2x^2-19x+6\phantom{)}}}\\
\underline{-~\phantom{(}(3x^3-6x^2)\phantom{-b)}}\\
0+8x^2-19x+6\phantom{)}\\
\underline{-~\phantom{()}(8x^2-16x)}\\
0-3x+6\phantom{)}\\
\underline{-~\phantom{()}(-3x+6)}\\
0\phantom{)}
\end{array}
\[3x^2+8x-3\\=3x^2+9x-x-3\\=\;3x(x+3)-1(x+3)\\=\;(3x-1)(x+3)\\Therefore,\;f(x)=(x-2)(x+3)(3x-1)\\\\\;\]
Q19. Find the value of ‘k’ if (x – 2) is a factor of x3 + 2x2 – kx + 10? [3] [2011]
Answer: k = 13
Step-by-step Explanation:
\[f(x)=\;\;x^3\;+\;2x^2\;–\;kx\;+\;10\\(x-2)\;is\;a\;factor\;of\;f(x).\\Therefore,\;f(2)=0\\{(2)}^3+2\times{(2)}^2-k\times2+10=0\\8+8-2k+10=0\\-2k+26=0\\-2k=-26\\k=13\]
Q20. When divided by x – 3 the polynomials x3 – px2 + x + 6 and 2x3 – x2 – (p + 3) x – 6 leave the same remainder. Find the value of ‘p’. [3] [2010]
Answer: p = 1
Step-by-step Explanation:
\[p(x)=\;x^3\;–\;px^2\;+\;x\;+\;6\;and\\q(x)=\;2x^3\;–\;x^2\;–\;(p\;+\;3)\;x\;–\;6\;\\when\;(x-3)\;divides\;p(x)\;and\;q(x),\;the\;remainders\;are\;same.\\Therefore,\;p(3)=q(3)\\{(3)}^3-p\times{(3)}^2+3+6={2\times(3)}^3-{(3)}^2-(p+3)\times3-6\\27-9p+9=54-9-3p-9-6\\-9p+3p+36=54-24\\-6p=30-36\\-6p=-6\\p=1\]
Q21. Use the Remainder Theorem to factorise the following expression:
2x3 + x2 – 13x + 6 [3] [2010]
Answer: (x-2)(x+3)(2x-1)
Step-by-step Explanation:
\[f(x)=\;2x^3\;+x^2-13x+6\;\\taking\;x=2,\;we\;have,\\f(2)=2\times{(2)}^3+{(2)}^2-13\times2+6\\\;\;\;\;\;\;=\;16+4-26+6\\\;\;\;\;\;\;=\;26-26\\\;\;\;\;\;\;=\;0\\Therefore,\;(x-2)\;is\;a\;factor\;of\;f(x).\\dividing\;f(x)\;by\;(x-2)\;we\;have,\\\\\]
\begin{array}{r}
2x^2+5x-3\phantom{)} \\
x-2{\overline{\smash{\big)}\,2x^3+x^2-13x+6\phantom{)}}}\\
\underline{-~\phantom{(}(2x^3-4x^2)\phantom{-b)}}\\
0+5x^2-13x+6\phantom{)}\\
\underline{-~\phantom{()}(5x^2-10x)}\\
0-3x+6\phantom{)}\\
\underline{-~\phantom{()}(-3x+6)}\\
0\phantom{)}
\end{array}
\[2x^2+5x-3\\=2x^2\;+6x-x-3\\=2x(x+3)-1(x+3)\\=\;(x+3)(2x-1)\\Therefore,\;f(x)=\;(x-2)(x+3)(2x-1)\\\\\;\;\;\;\\\\\]
Subscribe to my YouTube channel for more such educational content.
Sharing is caring!
