please help me this question is a not solve from me class 9th chapter name polynomial
Answers
Hope this will help you
Answer:
(i) 12x^2–7x+1
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = -7 and product =1×12 = 12
We get -3 and -4 as the numbers [-3+-4=-7 and -3×-4 = 12]
12x^2–7x+1
= 12x^2-4x-3x+1
= 4x(3x-1)-1(3x-1)
= (4x-1)(3x-1)
(ii) 2x^2+7x+3
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = 7 and product = 2×3 = 6
We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 6]
2x^2+7x+3
= 2x^2+6x+1x+3
= 2x (x+3)+1(x+3)
= (2x+1)(x+3)
(iii) 6x^2+5x-6
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = 5 and product = 6×-6 = -36
We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -36]
6x^2+5x-6
= 6x^2+9x–4x–6
= 3x(2x+3)–2(2x+3)
= (2x+3)(3x–2)
(iv) 3x^2–x–4
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = -1 and product = 3×-4 = -12
We get -4 and 3 as the numbers [-4+3 = -1 and -4×3 = -12]
3x^2–x–4
= 3x^2–x–4
= 3x^2–4x+3x–4
= x(3x–4)+1(3x–4)
= (3x–4)(x+1)
Question :
Factories :
i) 12x² - 7x + 1
ii) 2x² + 7x + 3
iii) 6x² + 5x - 6
iv) 3x² - x - 4
Answer :
i) 12x² - 7x + 1
→ 12x² - 3x - 4x + 1
→ 3x(4x - 1) - 1(4x - 1)
→ (3x - 1) (4x - 1) ....(Answer)
ii) 2x² + 7x + 3
→ 2x² + 6x + x + 3
→ 2x(x + 3) + 1(x + 3)
→ (2x + 1) (x + 3) ....(Answer)
iii) 6x² + 5x - 6
→ 6x² + 9x - 4x - 6
→ 3x(2x + 3) - 2(2x + 3)
→ (3x - 2) (2x + 3) .....(Answer)
iv) 3x² - x - 4
→ 3x² + 3x - 4x - 4
→ 3x(x + 1) - 4(x + 1)
→ (3x - 4) (x + 1) .....(Answer)
Note : We have factorise all the sum by middle term factor.
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Some Formulae for Factorisation :
- (a² - b²) = (a - b) (a + b)
- (a + b)² = (a² + 2ab + b²)
- (a - b)² = (a² - 2ab + b²)
- (a + b + c)² = (a² + b² + c²) + 2(ab + bc + ca)