Question

Subject: Factorization

1. Factorise:
(a) 14m5n4p² - 42m7n3p⁷ - 70m6n4p³
(b) 2a²(b² - c²) + b² (2c² - 2a²) + 2c²(a² - b²)

2. 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

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Answer 1

Solution:

1) Factorise:

a) 14m5n4p² - 42m7n3p⁷ - 70m6n4p³

In this expression 7mnp² is the common factor. This is explained as below: 14m5n4p² = 7 x 2 x m x 5 x n x 4 x p x p

- 42m7n3p⁷ = 7 x -6 x m x 7 x n x 3 x p x p x p x p x p x p x p

- 70m6n4p³ = 7 x -10 x m x 6 x n x 4 x p x p x p

The bolded factors are the common factors in each term and hence we have the common factor as 7mnp².

Taking out 7mnp² from each term, we are left with:

14m5n4p² ------ 40

- 42m7n3p⁷ ------ -126p⁵

-70m6n4p³ ------ -240p

So, the expression becomes:

7mnp² (40 - 126p⁵ - 240p)

Now we can factorise the terms in the brackets to find more common factors:

40 ------ 2 x 2 x 2 x 5

-126p⁵ ------ 2 x 3 x 3 x -7 x p x p x p x p x p

-240p ------ 2 x 2 x 2 x 2 x 3 x -5 x p

Here we can see that the only common factor is 2. Multiply that to 7mnp², we get 14mnp². Taking 14mnp² out of the expression:

= 14mnp² (20 - 63p⁵ - 120p)

b) 2a²(b² - c²) + b² (2c² - 2a²) + 2c²(a² - b²)

First of all, we need to open the brackets. Doing that, we get:

= 2a²b² - 2a²c² + 2b²c² - 2a²b² + 2a²c² - 2b²c²

Cancelling the like terms with different signs:

= 0

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

The algebraic identity a² - b² = (a + b)(a - b) is used here.

i) a² - 64

= a² - 8²

= (a + 8)(a - 8)

ii) 20a² - 45b²

= 5 [(4a²) - (9b²)]

= 5 [(2a + 3b) (2a - 3b)]

iii) 32x²y² - 8

= 2 (16x²y² - 4)

= 2 [(4xy)² - (2²)]

= 2 [(4xy + 2) (4xy - 2)]

iv) x² - 2xy + y² - z²

We know that (a - b)² = a² - 2ab + b²

So x² - 2xy + y²

= (x - y)²

Now,

(x - y)² - z²

= x² - y² - z²

= (x -  y + z) (x - y - z)

v) 49x² - 1

= (7x)² - 1²

= (7x + 1) (7x - 1)

Final Answers:

1. a) 14mnp² (20 - 63p⁵ - 120p)

b) 0

2. i) (a + 8)(a - 8)

ii) 5 [(2a + 3b) (2a - 3b)]

iii) 2 [(4xy + 2) (4xy - 2)]

iv) (x -  y + z) (x - y - z)

v) (7x + 1) (7x - 1)

Answer 2

Correct Question 1 :-

Factorise —

(a) 14m5n4p² - 42m7n3p⁷ - 70m6n4p³

(b) 2a²(b² - c²) + b² (2c² - 2a²) + 2c²(a² - b²)

Correct Answer 1 :-

Equation 1 :

We can factorize the first equation (a) 14m5n4p² - 42m7n3p⁷ - 70m6n4p³ as follows;

$\sf{14m5n4 {p}^{2} - 42m7n3 {p}^{7} - 70m6n4 {p}^{3} }$

${ \to \sf{14\left(m_{5}n_{4}p^{2}-3m_{7}n_{3}p^{7}-5m_{6}n_{4}p^{3}\right) }}$

${ \to \sf{p^{2}\left(m_{5}n_{4}-3m_{7}n_{3}p^{5}-5m_{6}n_{4}p\right) }}$

${ \to { \pmb{ \sf{ \red{14p^{2}\left(m_{5}n_{4}-3m_{7}n_{3}p^{5}-5m_{6}n_{4}p\right) }}}}}$

∴ Hence, the factorisation for $\sf{14m5n4 {p}^{2} - 42m7n3 {p}^{7} - 70m6n4 {p}^{3}}$ will be $ \sf{14p^{2}\left(m_{5}n_{4}-3m_{7}n_{3}p^{5}-5m_{6}n_{4}p\right) }$.

Equation 2 :

Now, we can factorize the second equation (b) 2a²(b² - c²) + b² (2c² - 2a²) + 2c²(a² - b²) as follows;

$\sf{2 {a}^{2} ( {b}^{2} - {c}^{2} ) + {b}^{2} (2 {c}^{2} - 2 {a}^{2} ) + 2 {c}^{2} ( {a}^{2} - {b}^{2} )}$

${ \to \sf{2a^{2}b^{2}-2a^{2}c^{2}+b^{2}\left(2c^{2}-2a^{2}\right)+2c^{2}\left(a^{2}-b^{2}\right) }}$

${ \to \sf{2a^{2}b^{2}-2a^{2}c^{2}+2b^{2}c^{2}-2b^{2}a^{2}+2c^{2}\left(a^{2}-b^{2}\right) }}$

${ \to \sf{-2a^{2}c^{2}+2b^{2}c^{2}+2c^{2}\left(a^{2}-b^{2}\right) }}$

${ \to \sf{-2a^{2}c^{2}+2b^{2}c^{2}+2c^{2}a^{2}-2b^{2}c^{2} }}$

${ \to \sf{2b^{2}c^{2}-2b^{2}c^{2} }}$

$\to { \pmb{ \sf{ \red{0 }}}}$

∴ Hence, the value for $\sf{2 {a}^{2} ( {b}^{2} - {c}^{2} ) + {b}^{2} (2 {c}^{2} - 2 {a}^{2} ) + 2 {c}^{2} ( {a}^{2} - {b}^{2} )}$ will be 0.

@TheOaky