Question

write answers pls fast it's long so I give 10 points ​

Answer 1

$\huge\mathfrak\color {teal}☆AnSwEr☆$

$\sf\orange {\underline {\sf Que.1\:Solution:-}}$

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

》Apply this identity $( {a}^{2} + {b}^{2} + {c}^{2} ) + 2xy + 2yz + 2zx = ( {a + b + c)}^{2}$

So,

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

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

Therefore, ${4x}^{2}+{9y}^{2}+{16z}^{2}+12xy-24yz-16xz={2x+3y-4z}^{2}$

$\sf\blue {\underline {\sf Que.2\:Solution:-}}$

${27x}^{3} + {8y}^{3}$

》Apply this identity $( {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )$

$( {3x)}^{3} + ( {2y)}^{3}$

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

$(3x + 2y)( {9x}^{2} - 6xy + {4y)}^{2}$

$\sf\pink{\underline {\sf Que.3\:Solution:-}}$

(a)

$12 {x}^{2} - 7x + 1$

$12 {x}^{2} - 4x - 3x + 1$

$4x(3x - 1) - (3x - 1)$

$(3x - 1)(4x - 1)$

(b)

$\sf\purple {\underline {\sf Que.4\:Solution:-}}$

$3 {x}^{2} - x - 4$

$3 {x}^{2} + 3x - 4x - 4$

$3x( x + 1) - 4(x + 1)$

$(x + 1)(3x - 4)$

$\sf\red{\underline {\sf Que.4\:Solution:-}}$

Let,

Divisor = 0

x+2 = 0

x = -2

$p(x) = {x}^{3} + 3 {x}^{2} + 3x + 1$

Putting, x = -2

$p( - 2) = {( - 2)}^{3} + 3 {( - 2)}^{2} + 3( - 2) + 1$

$= - 8 + 6 - 6 + 1$

$= - 8 - 0 + 1$

$= - 7$

Answer 2

Answer:

OKAY MATE.................... ❤