Question

answer the following questions step by step -

1)
$x^{2} - 2 {}^{2} - 8$
2)
$6x {}^{2} - 7x - 3$
3)
$4u {}^{2} + 8u$
4)
$3x {}^{2} - x - 4$
please answer the question properly spm will be deleted

Answer 1

(1)

Given Equation is x^2 - 2^2 - 8

= > x^2 - 12

$= \ \textgreater \ x^2 - ( \sqrt{12} )^2$

$= \ \textgreater \ (x + \sqrt{12} )(x - \sqrt{12} )$

$= \ \textgreater \ (x + 2 \sqrt{3} )(x - 2 \sqrt{3} )$



(2) 

Given 6x^2 - 7x - 3

= > 6x^2 + 2x - 9x - 3

= > 2x(3x + 1) - 3(3x + 1)

= > (2x - 3)(3x + 1).


(3)

Given 4u^2 + 8u

= > 4u(u + 2)


(4)

Given 3x^2 - x - 4

= > 3x^2 + 3x - 4x - 4

= > 3x(x + 1) - 4(x + 1)

= > (3x - 4)(x + 1).



Hope this helps!

Answer 2

Hey

Here is your answer,

_____________________________
1) x^2-2x-8=0
X^2 +2x -4x -8=0
X(x+2)-4(x+2)=0
(X+2)(x-4)=0
X+2=0
X=-2
X-4=0
X=4

_____________________________
2) 6x^2 -7x -3=0
6x^2 +2x-9x-3=0
2x(3x+1) - 3(3x+1)=0
(3x+1)(2x -3)=0
3x+1=0
X=-1/3
2x-3=0
X=3/2

_____________________________
3)4u^2-8u=0
4u(u-2)=0
4u=0
U=0
U-2=0
u=2

_____________________________
4)3x^2 -x-4=0
3x^2+3x -4x -4=0
3x(x+1)-4(x+1)=0
(3x -4)(x+1)=0
3x-4=0
X=4/3
X+1=0
X=-1

_____________________________

Hope it helps you!
Please mark it as brainliest answer!