Question

factorise the following



class 9th chapter 2 exercise 2.5​

Answer 1

Answer:

(i) (2x+3y-4z)²

(ii) (√2x-y-√8z)²

Answer 2

$\huge \fbox \green{❤Answer:}$

We Have,

$\sf \colorbox{skyblue} {Ans. (i)}$

$4 {x}^{2} + 9 {y}^{2} + 16 {z}^{2} + 12xy - 24yz - 16xz$

$= (2x {)}^{2} + (3y {)}^{2} + ( - 4z {)}^{2} + 2 \times 2x \times 3y + 2 \times 3y \times ( - 4z) + 2 \times ( - 4z) \times 2x$

$[∴ {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = (a + b + c {)}^{2}]$

$= (2x + 3y - 4z {)}^{2}$

$= (2x + 3y - 4z)(2x + 3y - 4z)$

$\sf \colorbox{pink} {Ans. (ii)}$

$2 {x}^{2} + {y}^{2} + 8 {z}^{2} - 2 \sqrt{2xy} + 4 \sqrt{2yz} - 8xz$

$= ( - \sqrt{2x} {)}^{2} + (y {)}^{2} + (2 \sqrt{2z} {)}^{2} + 2 \times ( - \sqrt{2x} ) \times y + 2 \times y \times 2 \sqrt{2z} + 2 \times 2 \sqrt{2z} \times ( - \sqrt{2x} )$

$[∴ {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = (a + b + c {)}^{2}]$

$= ( - \sqrt{2x} + y + 2 \sqrt{2z} {)}^{2}$

$= ( - \sqrt{2x} + y + 2 \sqrt{2z}) ( - \sqrt{2x} + y + 2 \sqrt{2z}$

$\\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}$