Question

heya

please solve this question it's from factorisation chapter of STD 8.


4n^2-8n+3

Answer 1

$4 {n}^{2} - 8n + 3 \\ = 4 {n}^{2} - (6 + 2)n + 3 \\ = 4 {n}^{2} - 6n - 2n + 3 \\ = 2n(2n - 3) - 1(2n - 3) \\ = (2n - 3)(2n - 1)$
[Note: we have multiplied 4 (n²) and 3 (constant). After that we have done prime factorisation of the product ( i.e. 12) and includes the prime factorisation number in getting the number 8 (n).]

Hope it'll help..:)

Answer 2

Heya !

Here is your answer

Question -
$4n^{2} - 8n + 3$
This question can be solved by splitting the middle term


Answer - 4 × 3 = 12

Now we have to find factors of 12 such that there addition is - 8 and multiplication 12

thus the numbers are

6 and 2

- 6 - 2 = -8

-6 × -2 = 12

thus,

${4n}^{2} - 2n - 6n +3$

thus,

taking common

2n( 2n - 1) - 3(2n -1)

therefore,

(2n - 3) (2n- 1)

Hope it helps !