Class 10 Mathematics Chapter 7, Factorization is a crucial topic in ICSE Class 10 mathematics. It involves breaking down algebraic expressions or numbers into their constituent factors. Understanding factorization is essential as it forms the foundation for many other mathematical concepts. Here are some factorisation class 10 ICSE important questions.
Factorization as we study in factorisation class 10 ICSE is the process of expressing a number or algebraic expression as the product of its factors. Factors are numbers or expressions that multiply together to yield the original number or expression. The main goal of factorization is to simplify complex expressions, solve equations, and understand the underlying structure of mathematical concepts.
Common Factorization Techniques:
Common Factors: Identifying and factoring out the common factors shared by multiple terms in an expression.
Example: Factorize 6x + 9y. The common factor is 3, so the expression becomes 3(2x + 3y).
Difference of Squares: Recognizing expressions in the form of a2 - b2 and factoring them as (a + b)(a - b).
Example: Factorize x2 - 9. The expression factors as (x + 3)(x - 3).
In ICSE Class 10 mathematics, factorization refers to the process of expressing an algebraic expression or a number as the product of its constituent factors. Factors are numbers or algebraic expressions that, when multiplied together, result in the original expression or number. Factorization is an essential skill in mathematics and is used to simplify expressions, solve equations, and understand the underlying structure of mathematical concepts.
For example, consider the algebraic expression 2x2 + 4x. Factorization of this expression involves finding two expressions that, when multiplied, result in 2x2 + 4x. In this case, the factors are 2x and (x + 2), so the factorization is:
2x2 + 4x = 2x(x + 2).
Factorization is also used to break down numbers into their prime factors. For instance, the prime factorization of the number 12 is:
12 = 2 * 2 * 3.
Ans. (b) 52
Explanation:
Let
p(x) = 6x3+ x2 – 2x + 4
When, p(x) is divided by x – 2,
Remainder = p(x = 2)
= 6(2)3 + (2)2– 2(2) + 4
= 48 + 4 – 4 + 4 = 52
Ans. (d) 16
Explanation:
Let f(x) = 3x3– x2 – px – 4 ...(i)
Since, (x + 2) is a factor of f(x), f(–2) = 0
⇒ 3(–2)3 – (–2)3 – p(–2) – 4 = 0
⇒ –24 – 4 + 2p – 4 = 0
⇒ 2p = 32
⇒ p = 16
Explanation:
Given expression is 2x3 – x2 – px – 2 and x – 2 is the factor.
(i) x – 2 = 0, x = 2 in expression
2 (2)3 – (2)2 – p (2) – 2 = 0
⇒ 16 – 4 – 2p – 2 = 0
⇒ 10 – 2p = 0
⇒ p = 5
(ii) Putting the value of p
∴ 2x3 – x2 – 5x – 2 = (x – 2) (2x2 + 3x + 1)
The expression can be the written as
(2x3 + x2 + 1) (x – 2) or (2x + 1) (x + 1) (x – 2).
Explanation:
(i) For p (x), we have to find all possible factors of the constant -6
(By factor theorem, we have:)
Factors of - 6 are ∓1, ∓2, ∓3, ∓6, etc. and on putting the value in p (x) we get:
By hit and trial: for x = 1 we get p(i) = 13 - 6(1)2 +11(1)-6 = 0.
Which clearly shows that (x-1) is a factor of p (x).
∴ on dividing p(x) by (x-1)will give:
Clearly p(x) = (x - 1)(x2 - 5x + 6) which on further factorisation gives:
p(x)=(x - 1)(x - 2)(x - 3)...(A)
Similarly solving for q(x) we have:
q(x)=(x - 2)(x - 3)(x - 1)...(B)
∴ L.C.M. = (x - 1)(x - 2)(x - 3)(x + 1)...(c)
Explanation:
Let P (x) = 2x3 + ax2 + bx – 2 when P(x) is divided by 2x – 3
P\(\begin{pmatrix} \frac{3}{2} \end{pmatrix}\)=2 \(\begin{pmatrix} \frac{3}{2} \end{pmatrix}^3\)+a\(\begin{pmatrix} \frac{3}{2} \end{pmatrix}^2\)+b\(\begin{pmatrix} \frac{3}{2} \end{pmatrix}\)-2=7
=\(\frac{27}{4}\) + \(\frac{9}{4}\) a + \(\frac{3}{2}\) b - 2 = 7
= 9a + 6b = 28 + 8 – 27
= 9a + 6b = 9
= 3a + 2b = 3 …(i)
Similarly when P(x) is divided by x + 2
⇒ x = – 2
∴ 2(– 2)3 + a(– 2)2 + b(– 2) – 2 = 0
⇒ – 16 + 4a – 2b – 2 = 0
⇒ 4a – 2b = 18 …(ii)
On solving equations (i) and (ii)
3a + 2b = 3
\(\underline{4a – 2b = 18}\) (On adding (ii)
7a = 21
a = 3
On substituting value of a in equation (i)
3 × 3 + 2b = 3
2b = 3 – 9
b=\(\frac{-6}{2}\)=-3
b = – 3
a = 3, b = – 3
On substituting value of a and b
2x3 + 3x2 – 3x – 2
When x + 2 is a factor
2x2 – x – 1 = 2x2 – 2x + x – 1
= 2x(x – 1) + 1 (x – 1)
= (x – 1) (2x + 1)
Hence, required factors are
(x – 1) (x + 2) (2x + 1)
The chapter factorization class 10 ICSE is a foundational and pivotal topic that serves as a cornerstone for various mathematical concepts and practical applications. Mastering these factorisation class 10 ICSE important questions equips you with valuable problem-solving skills for everyday situations, making it a crucial part of your math education.
Ans: Factorization is the process of expressing an algebraic expression or a number as the product of its constituent factors. It is crucial in mathematics for simplifying expressions, solving equations, and understanding mathematical relationships.
Ans: Prime factors are the smallest prime numbers that multiply together to give a given number. To find them, you can use techniques like prime factorization or factor tree diagrams.
Ans: Algebraic factorization involves breaking down algebraic expressions into their constituent factors, while numerical factorization involves finding the prime factors of a number.
Ans: You can factorize quadratic trinomials using techniques like the product-sum method or trial and error to find two binomials that multiply to the trinomial.
Ans: A perfect square trinomial is the square of a binomial expression. It is factorized as the square of the binomial.