Factor the following: i. 9a² + b² + 4c² - 6ab +12ac - 4bc ii. 16s² + 25t² - 40st Class 9
Factor the following: i. 9a² + b² + 4c² - 6ab +12ac - 4bc ii. 16s² + 25t² - 40st Class 9
Question 1.
Factor the following: Class 9
i. 9a² + b² + 4c² - 6ab +12ac - 4bc
ii. 16s² + 25t² - 40st
iii. r² - r - 42
iv. 49g² + 14gh + h²
v. 64u² + 121u² + 4w² - 176uv - 32uw + 44vw
Solution:
i. 9a² + b² + 4c² - 6ab + 12ac - 4bc
= 9a² + b² + 4c² - 6ab - 4bc + 12ac
Writing in the form
A² + B² + C² + 2AB + 2BC + 2CA, we get
9a² + b² + 4c² - 6ab + 12ac - 4bc
= (3a)² + (-b)² + (2c)² + 2(3a)(-b) + 2(-b)(2c) + 2(2c)(3a),
where A = 3a, B = - b, C = 2c
∴ 9a² + b² + 4c² - 6ab + 12ac - 4bc
= [3a + (-b) + 2c]² ...[∵ (A + B + C)²
= A² + B² + C² + 2AB + 2BC + 2CA]
= (3a - b + 2c)²
= (3a - b + 2c)(3a - b + 2c)
ii. 16s² + 25t² - 40st = 16s² - 40st + 25t²
Writing in the form a² + 2ab + b², we get
16s² + 25t² - 40st = (4s)² - 2(4s) (5t) + (5t)²,
where a = 4s, b = 5f
∴ 16s² + 25t² - 40st
= (4s - 5t)² .....[∵ (a-b)² = a² - 2ab + b²]
= (4s - 5t)(4s - 5t)
iii. r² - r - 42
Comparing
r² - r - 42 with r² + (a + b)r + ab, we get
a + b = -1 and ab = - 42
Since (- 7) + 6 = -1 and (- 7) × (6) = - 42,
a = -7 and b = 6 or vice-versa
∴ r² - r - 42 = (r - 7)(r + 6)
iv. 49g² + 14gh + h²
Writing in the form a² + 2ab + b², we get
49g² + 14gh + h² = (7g)² + 2(7g)(h) + h²,
where a = 7g, b = h
∴ 49g² + 14gh + h²
= (7g + h)² ... [∵ (a + b)² = a² + 2ab + b²]
= (7g + h)(7g + h)
v. 64u² + 121v² + 4w² - 176uv - 32uw + 44vw
= 64u² + 121v²+ 4w² - 176uv + 44vw - 32uw
Writing in the form
a² + b² + c² + 2ab + 2bc + 2ca, we get
64u² + 121v² + 4w² - 176uv - 32uw + 44vw
= (8u)² + (-11v)² + (-2w)² + 2(8u)(-11v) + 2(-11v)(-2w) + 2(8u)(-2w),
where a = 8u, b = -11v, C = -2w
∴ 64u² + 121M2 + 4w2 - 176uv - 32uw + 44vw
= (8u - 11v - 2w)² ... [∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]
= (8u - 11v - 2w)(8u - 11v - 2w)