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Exercise 14.3 of Chapter 14 (Factorisation) in the Class 8 NCERT Mathematics textbook focuses on factoring algebraic expressions using various methods.
This exercise helps students understand how to break down complex expressions into simpler factors, which is a crucial skill in algebra.
Solution:
Given: 4(x - 5) = 4x - 5
Now we check if the given statement is correct or not
Taking L.H.S:
= 4(x - 5)
= 4x - 20
L.H.S ≠ R.H.S
So, the correct solution of 4(x - 5) = 4x - 20
Solution:
Given: x(3x + 2) = 3x2+ 2
Now we check if the given statement is correct or not
Taking L.H.S:
= x(3x + 2)
= 3x2 + 2x
L.H.S ≠ R.H.S
So, the correct solution of x(3x + 2) = 3x2+ 2x
Solution:
Given: 2x + 3y = 5xy
Here, L.H.S ≠ R.H.S
So, the correct solution of 2x + 3y = 2x + 3y
Solution:
Given: x + 2x + 3x = 5x
Now we check if the given statement is correct or not
Taking L.H.S:
= x + 2x + 3x
= 6x
L.H.S ≠ R.H.S
So, the correct solution of x + 2x + 3x = 6x
Solution:
Given: 5y + 2y + y – 7y = 0
Now we check if the given statement is correct or not
Taking L.H.S:
= 5y + 2y + y – 7y
= y
L.H.S ≠ R.H.S
So, the correct solution of 5y + 2y + y – 7y = y
Solution:
Given: 3x + 2x = 5x2
Now we check if the given statement is correct or not
Taking L.H.S:
= 3x + 2x
= 5x
L.H.S ≠ R.H.S
So, the correct solution of 3x + 2x = 5x
Solution:
Given: (2x)2+ 4(2x) + 7 = 2x2+ 8x + 7
Now we check if the given statement is correct or not
Taking L.H.S:
= (2x)2+ 4(2x) + 7
= 4x2 + 8x + 7
L.H.S ≠ R.H.S
So, the correct solution of (2x)2+ 4(2x) + 7 = 4x2 + 8x + 7
Solution:
Given: (2x)2+ 5x = 4x + 5x = 9x
Here,
4x + 5x ≠ 9x
So, we check (2x)2+ 5x = 4x + 5x or not
Taking L.H.S:
= (2x)2+ 5x
= 4x2 + 5x
L.H.S ≠ R.H.S
So, the correct solution of (2x)2+ 5x = 4x2 + 5x
Solution:
Given: (3x + 2)2= 3x2+ 6x + 4
Now we check if the given statement is correct or not
Taking L.H.S:
= (3x + 2)2
= 9x2 + 6x + 4
L.H.S ≠ R.H.S
So, the correct solution of (3x + 2)2 = 9x2 + 6x + 4
Solution:
Given: x2+ 5x + 4
Now substitute the value of x = -3 in the given equation,
= (-3)2+ 5(-3) + 4
= 9 - 15 + 4
= -2
So the correct solution of x2+ 5x + 4 = -2
Solution:
Given: x2 - 5x + 4
Now substitute the value of x = -3 in the given equation,
= (-3)2 - 5(-3) + 4
= 9 + 15 + 4
= 28
So the correct solution of x2 - 5x + 4 = 28
Solution:
Given: x2 + 5x
Now substitute the value of x = -3 in the given equation,
= (-3)2 + 5(-3)
= 9 - 15
= -6
So the correct solution of x2 + 5x = -6
Solution:
Given: (y - 3)2 = y2 - 9
Now we check if the given statement is correct or not
Taking L.H.S:
= (y - 3)2
= y2 - 6y + 9
L.H.S ≠ R.H.S
So, the correct solution of (y - 3)2 = y2 - 6y + 9
Solution:
Given: (z + 5)2 = z2+ 25
Now we check if the given statement is correct or not
Taking L.H.S:
= (z + 5)2
= z2 + 10z + 25
L.H.S ≠ R.H.S
So, the correct solution of (z + 5)2 = z2 + 10z + 25
Solution:
Given: (2a + 3b) (a - b) = 2a2 - 3b2
Now we check if the given statement is correct or not
Taking L.H.S:
= (2a + 3b) (a - b)
= 2a2 - 2ab + 3ab - 3b2
= 2a2 - 3b2 + ab
L.H.S ≠ R.H.S
So, the correct solution of (2a + 3b) (a - b) = 2a2 - 3b2 + ab
Solution:
Given: (a + 4) (a + 2) = a2 + 8
Now we check if the given statement is correct or not
Taking L.H.S:
= (a + 4) (a + 2)
= a2 + 2a + 4a + 8
= a2 + 6a + 8
L.H.S ≠ R.H.S
So, the correct solution of (a + 4) (a + 2) = a2 + 6a + 8
Solution:
Given: (a - 4) (a - 2) = a2 - 8
Now we check if the given statement is correct or not
Taking L.H.S:
= (a - 4) (a - 2)
= a2 - 2a - 4a + 8
= a2 - 6a + 8
L.H.S ≠ R.H.S
So, the correct solution of (a - 4) (a - 2) = a2 - 6a + 8
Solution:
Given: 3x2/3x2 = 0
Now we check if the given statement is correct or not
Taking L.H.S:
= 3x2/3x2
= 1
L.H.S ≠ R.H.S
So, the correct solution of 3x2/3x2 = 1
Solution:
Given: (3x2 + 1)/(3x2) = 1 + 1 = 2
Now we check if the given statement is correct or not
Taking L.H.S:
= (3x2 + 1)/(3x2)
= 3x2/(3x2)+ 1/(3x2)
= 1 + 1/(3x2)
L.H.S ≠ R.H.S
So, the correct solution of (3x2 + 1)/(3x2) = 1 + 1/(3x2)
Solution:
Given: (3x)/(3x + 2) = 1/2
Here, L.H.S ≠ R.H.S
So, the correct solution of (3x)/(3x + 2) = (3x)/(3x + 2)
Solution:
Given: 3/(4x + 3) = 1/4x
Here, L.H.S ≠ R.H.S
So, the correct solution of 3/(4x + 3) = 3/(4x + 3)
Solution:
Given: (4x + 5)/(4x) = 5
Now we check if the given statement is correct or not
Taking L.H.S:
= (4x + 5)/(4x)
= (4x/4x) + 5/4x)
= 1 + 5/4x
L.H.S ≠ R.H.S
So, the correct solution of (4x + 5)/(4x) = 1 + 5/4x
Solution:
Given: (7x + 5)/5 = 7x
Now we check if the given statement is correct or not
Taking L.H.S:
= (7x + 5)/5
= (7x/5) + 5/5)
= 7x/5 + 1
L.H.S ≠ R.H.S
So, the correct solution of (7x + 5)/5 = 7x/5 + 1
Exercise 14.4 of Chapter 14 on Factorisation introduces students to the crucial algebraic skill of division. This exercise builds upon previously learned factorization techniques and extends them to the realm of dividing polynomials by monomials and binomials. Through a series of carefully crafted problems, students learn to apply the distributive property, identify common factors, and use long division for more complex polynomial divisions. The problems range from simple divisions of monomials to more challenging divisions of quadratic expressions by linear factors. This exercise not only reinforces students' understanding of algebraic manipulation but also prepares them for more advanced topics in algebra, such as polynomial long division, synthetic division, and the factor theorem. Mastery of these division techniques is essential for simplifying rational expressions, solving higher-degree equations, and understanding the behavior of polynomial functions in higher mathematics. By working through these problems, students develop their analytical thinking, pattern recognition, and problem-solving skills, all of which are fundamental to success in mathematics and related fields.`
