Factorise: x² - 10x + 21
Solution:
Given, x² - 10x + 21
The process of finding two or more expressions whose product is the given expression is known as the factorization of algebraic expressions.
This expression is solved by splitting the middle term ⇒ Split the middle term (Term with x) in such a way that the factors of the product of the coefficient of the first term (x²) and the third term (constant term) add up to make the second term (-10x in this case) and see that the factor pair satisfies as the sum, equal to the middle/second term.
Here, first term = x², second/middle term = -10x and third term = 21
The product of the first and the third term = 1 × 21 = 21
The factor pairs of 21 are : (1, 21) and (3, 7).
Here, we can see that the factor pair (3, 7) satisfies our purpose, but middle term is negative hence the factor pair (-3, -7) is taken,
-3 - 7 = -10 and -3 × -7 = 21
Hence, the expression can be written as:
x² - 3x - 7x + 21
= x(x - 3) - 7(x - 3)
Taking out the common term (x - 3)
= (x - 3)(x - 7)
✦ Try This: Factorise: p² - 25p + 126
Given, p² - 25p + 126
= p² - 18p - 7p + 126
= p(p - 18) - 7(p - 18)
= (p - 7)(p - 18)
☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 9
NCERT Exemplar Class 8 Maths Chapter 7 Problem 91(viii)
Factorise: x² - 10x + 21
Summary:
Factorising x² - 10x + 21 we get, (x - 3)(x - 7)
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