Chapter 7: Algebraic Expressions and Identities & Factorisation

Solution:-

(a) (p – 19) (p + 2)

Factorised form of p² – 17p – 38 is = p² – 19p + 2p – 38

Take out the common factors,

= p (p – 19) + 2 (p – 19)

Again take out the common factor,

= (p – 19) (p + 2)

Solution:-

(c) ½pr

On dividing 57p²qr by 114pq,

It can be expanded as = (57 × p × p × q × r)/(114 × p × q)

= 57pr/114 … [divide both numerator and denominator by 57]

= ½pr

Solution:-

(c) p + 2

On dividing p (4p² – 16) by 4p (p – 2)

= (p((2p)² – (4)²))/ (4p(p – 2))

= ((2p – 4) × (2p + 4))/(4(p – 2))

Take out the common factors

= ((2(p – 2)) × (2 (p + 4)))/(4(p -2))

= (4(p – 2)(p + 2))/ (4(p – 2))

= p + 2

Solution:-

(a) 1

Considering the two monomials 3ab and 2cd, there is no common factor except 1.

Solution:-

(c) x

An irreducible factor is a factor which cannot be expressed further as a product of factors. Such a factorisation is called an irreducible factorisation.

24x²y² = 2 × 2 × 2 × 3 × x × x × y × y

Therefore an irreducible factor is x.

Solution:-

(c) 2

Number of factors of (a + b)² is = (a + b) (a + b) no further factorisation is possible.

Solution:-

(b) 3 (x – 8)

The factorised form of 3x – 24 is,

Take out 3 as common,

= 3 (x – 8)

Solution:-

(b) (x + 2), (x – 2)

The factors of x² – 4 are,

X² – 4 = x² – 2²

= (x + 2) (x – 2)

Solution:-

(d) 3x

The value of (– 27x²y) ÷ (– 9xy) = (-27 × x × x × y)/(- 9 × x × y)

= (27/9)x … [divide both numerator and denominator by 3]

= 3x

Solution:-

(b) x² + 2

The value of (2x² + 4) ÷ 2 = (2x² + 4)/2

= (2(x² + 2))/2

= x² + 2