FACTORISATION IMPORTANT CLASS 8

 FACTORISATION IMPORTANT

BY AARISH SIR

1. Express the following as in the form of (a+b)(a-b)

(i) a² – 64

(ii) 20a² – 45b²

(iii) 32x²y² – 8

(iv) x² – 2xy + y² – z²

(v) 49x² – 1

Solution:

For representing the expressions in (a+b)(a-b) form, use the following formula

a² – b² = (a+b)(a-b)

(i) a² – 64 = a² – 8² = (a + 8)(a – 8)

(ii) 20a² – 45b² = 5(4a² – 9b²) = 5(2a + 3b)(2a – 3b)

(iii) 32x²y² – 8 = 8( 4x²y² – 1) = 8(2xy + 1)(2xy – 1)

(iv) x² – 2xy + y² – z² = (x – y)² – z² = (x – y – z)(x – y + z)

(v) 49x² – 1 = (7x)² – (1)² = (7x + 1)(7x – 1)

2. Verify whether the following equations are correct. Rewrite the incorrect equations correctly.

(i) (a + 6)² = a² + 12a + 36

(ii) (2a)² + 5a = 4a + 5a

Solution:

(i) (a + 6)² = a² + 12a + 36

Here, LHS = (a + 6)² = a² + 12a + 36

Now, RHS = a² + 12a + 36

Hence, LHS = RHS.

(ii) (2a)² + 5a = 4a + 5a

Here, LHS = (2a)² + 5a = 4a² + 5a

Now, RHS = 4a + 5a

So, LHS ≠ RHS

Correct equation: (2a)² + 5a = 4a² + 5a

3. For a = 3, simplify a² + 5a + 4 and a² – 5a

Solution:

Substitute the value of a = 3 in the given equations.

a² + 5a + 4 = 3² + 5(3) + 4 = 9 + 15 + 4 = 28

And,

a² – 5a = 3² – 5(3) = 9 – 15 = -6

Long Answer Type Questions:

4. Find the common factors of the following:

(i) 6 xyz, 24 xy² and 12 x²y

(ii) 3x² y³, 10x³ y² and 6x² y² z

Solution:

(i) 6 xyz = 2 × 3 × x × y × z

24 xy² = 2 × 2 × 2 × 3 × x × y × y

12 x²y = 2 × 2 × 3 × x × x × y

Thus, the common factors are common factors of 6 xyz, 24 xy² and 12 x²y are 2, 3, x, y and, (2 × 3 × x × y) = 6xy

(ii) 3x² y³ = 3 × x × x × y × y × y

10x³ y² = 2 × 5 × x × x × x × y × y

6 x² y² z = 3 × 2 × x × x × y × y × z

Now, the common factors of 3x² y³, 10x³ y² and 6x² y² z are x², y² and, (x² × y²) = x² y²

5. Factorize the following expressions:

(i) 54x³y + 81x⁴y²

(ii) 14(3x – 5y)³ + 7(3x – 5y)²

(iii) 15xy + 15 + 9y + 25x

Solution:

(i) 54x³y + 81x⁴y²

= 2 × 3 × 3 × 3 × x × x × x × y + 3 × 3 × 3 × 3 × x × x × x × x × y × y

= 3 × 3 × 3 × x × x × x × y × (2 + 3 xy)

= 27x³y (2 + 3 xy)

(ii) 14(3x – 5y)³ + 7(3x – 5y)²

= 7(3x – 5y)² [2(3x – 5y) +1]

= 7(3x – 5y)² (6x – 10y + 1)

(iii) 15xy + 15 + 9y + 25x

Rearrange the terms as:

15xy + 25x + 9y + 15

= 5x(3y + 5) + 3(3y + 5)

Or, (5x + 3)(3y + 5)

6. Factorize (x + y)² – 4xy

Solution:

To solve this expression, expand (x + y)²

Use the formula:

(x + y)² = x² + 2xy + y²

(x + y)² – 4xy = x² + 2xy + y² – 4xy

= x² + y² – 2xy

We know, (x – y)² = x² + y² – 2xy

So, factorization of (x + y)² – 4xy = (x – y)²

7. Factorize x² + 6x – 16

Solution:

To factorize, it should be checked that the sum of factors of 16 should be equal to 6.

Here, 16 = -2 × 8 and 8 + (-2) = 6

So,

x² + 6x – 16 = x² – 2x + 8x – 16

= x(x – 2) + 8(x – 2)

= (x + 8) (x – 2)

Hence, x² + 6x – 16 = (x + 8) (x – 2)

8. Solve for (4x² – 100) ÷ 6(x + 5)

Solution:

= ⅔ (x – 5)

TRY THESE ALSO

1. Factorise:
   (а) 14m5n4p2 – 42m7n3p7 – 70m6n4p3
   (b) 2a2(b2 – c2) + b2(2c2 – 2a2) + 2c2(a2 – b2)
2.  The area of a rectangle is 6a2 + 36a and 36a width. Find the length of the rectangle.
3. What are the common factors of the following terms?
   (a) 25x2y, 30xy2
   (b) 63m3n, 54mn4