Question 21 of 43intermediate🔧 ApplyShort Answer5 marks

Factor the following:

(i)

9a29a^{2} + b2b^{2} + 4c24c^{2} - 6ab + 12ac - 4bc

Answer

(3a - b + 2c)(3a - b + 2c)

(ii)

16s216s^{2} + 25t225t^{2} - 40st

Answer

(4s - 5t)(4s - 5t)

(iii)

r2r^{2} - r - 42

Answer

(r - 7)(r + 6)

(iv)

49g249g^{2} + 14gh + h2h^{2}

Answer

(7g + h)(7g + h)

(v)

64u264u^{2} + 121v2121v^{2} + 4w24w^{2} - 176uv - 32uw + 44vw

Answer

(8u - 11v - 2w)(8u - 11v - 2w)

Explanation

The textbook introduces several algebraic identities for factorization. For expressions with three squared terms and three cross terms, we use the identity (x + y + z)^2 = x2x^{2} + y2y^{2} + z2z^{2} + 2xy + 2xz + 2yz, carefully determining the signs of x, y, and z from the cross terms. For perfect square trinomials, we apply (a + b)^2 = a2a^{2} + 2ab + b2b^{2} or (a - b)^2 = a2a^{2} - 2ab + b2b^{2}. For quadratic trinomials of the form x2x^{2} + (a+b)x + ab, we find two numbers that multiply to the constant term and add to the coefficient of the middle term, factoring as (x + a)(x + b).