Factorise: y² + 4y - 21
Solution:
Given, y² + 4y - 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 y) in such a way that the factors of the product of the coefficient of the first term (y²) and the third term (constant term) add up to make the second term (4y in this case) and see that the factor pair satisfies as the sum, equal to the middle/second term.
Here, first term = y², second/middle term = 4y and third term = -21
The product of the coefficient of y² and the third term = 1 × (-21) = -21
The factor pairs of -21 are : (1, -21), (3, -7), (-3, 7) and (-1, 21).
Here, we can see that the factor pair (-3, 7) satisfies our purpose as,
-3 + 7 = 4 and -3 × 7 = -21
Hence, the expression can be written as:
y² - 3y + 7y - 21
= y(y - 3) + 7(y - 3)
Taking out the common term (y - 3)
= (y - 3)(y + 7)
✦ Try This: Factorise: x² + x - 20
Given, x² + x - 20
= x² - 4x + 5x - 20
= x(x - 4) + 5(x - 4)
= (x + 5)(x - 4)
☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 9
NCERT Exemplar Class 8 Maths Chapter 7 Problem 91(v)
Factorise: y² + 4y - 21
Summary:
Factorising y² + 4y - 21 we get, (y - 3)(y + 7)
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