find the hcf of the following polynomials(step-by-step explanations)(x²-4)(x²+5x+6) and (x+3)²(3x²+x-6) About the author Parker
Answer: Page No 220: Question 1: Find the common factors of the terms (i) 12x, 36 (ii) 2y, 22xy (iii) 14pq, 28p2q2 (iv) 2x, 3×2, 4 (v) 6abc, 24ab2, 12a2b (vi) 16×3, −4×2, 32x (vii) 10pq, 20qr, 30rp (viii) 3x2y3, 10x3y2, 6x2y2z ANSWER: (i) 12x = 2 × 2 × 3 × x 36 = 2 × 2 × 3 × 3 The common factors are 2, 2, 3. And, 2 × 2 × 3 = 12 (ii) 2y = 2 × y 22xy = 2 × 11 × x × y The common factors are 2, y. And, 2 × y = 2y (iii) 14pq = 2 × 7 × p × q 28p2q2 = 2 × 2 × 7 × p × p × q × q The common factors are 2, 7, p, q. And, 2 × 7 × p × q = 14pq (iv) 2x = 2 × x 3×2 = 3 × x × x 4 = 2 × 2 The common factor is 1. (v) 6abc = 2 × 3 × a × b × c 24ab2 = 2 × 2 × 2 × 3 × a × b × b 12a2b = 2 × 2 × 3 × a × a × b The common factors are 2, 3, a, b. And, 2 × 3 × a × b = 6ab (vi) 16×3 = 2 × 2 × 2 × 2 × x × x × x −4×2 = −1 × 2 × 2 × x × x 32x = 2 × 2 × 2 × 2 × 2 × x The common factors are 2, 2, x. And, 2 × 2 × x = 4x (vii) 10pq = 2 × 5 × p × q 20qr = 2 × 2 × 5 × q × r 30rp = 2 × 3 × 5 × r × p The common factors are 2, 5. And, 2 × 5 = 10 (viii) 3x2y3 = 3 × x × x × y × y × y 10x3y2 = 2 × 5 × x × x × x × y × y 6x2y2z = 2 × 3 × x × x × y × y × z The common factors are x, x, y, y. And, x × x × y × y = x2y2 Step-by-step explanation: plz thanks my answer and mark as brilliant. Reply
Answer:
Page No 220:
Question 1:
Find the common factors of the terms
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p2q2
(iv) 2x, 3×2, 4
(v) 6abc, 24ab2, 12a2b
(vi) 16×3, −4×2, 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x2y3, 10x3y2, 6x2y2z
ANSWER:
(i) 12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
The common factors are 2, 2, 3.
And, 2 × 2 × 3 = 12
(ii) 2y = 2 × y
22xy = 2 × 11 × x × y
The common factors are 2, y.
And, 2 × y = 2y
(iii) 14pq = 2 × 7 × p × q
28p2q2 = 2 × 2 × 7 × p × p × q × q
The common factors are 2, 7, p, q.
And, 2 × 7 × p × q = 14pq
(iv) 2x = 2 × x
3×2 = 3 × x × x
4 = 2 × 2
The common factor is 1.
(v) 6abc = 2 × 3 × a × b × c
24ab2 = 2 × 2 × 2 × 3 × a × b × b
12a2b = 2 × 2 × 3 × a × a × b
The common factors are 2, 3, a, b.
And, 2 × 3 × a × b = 6ab
(vi) 16×3 = 2 × 2 × 2 × 2 × x × x × x
−4×2 = −1 × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
The common factors are 2, 2, x.
And, 2 × 2 × x = 4x
(vii) 10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
The common factors are 2, 5.
And, 2 × 5 = 10
(viii) 3x2y3 = 3 × x × x × y × y × y
10x3y2 = 2 × 5 × x × x × x × y × y
6x2y2z = 2 × 3 × x × x × y × y × z
The common factors are x, x, y, y.
And,
x × x × y × y = x2y2
Step-by-step explanation:
plz thanks my answer and mark as brilliant.