How do you factorise and then solve x^2=16, x^2+5x+6=0, and 3x^2-10x-8=0?

Answer 1

1.x^2=16
=>x^2-16=0
(x)^2-(4)^2=0
(x+4)(x-4)=0
Therefore,either x+4=0 or x-4=0
=>x=+4 or -4
2.X^2+5x+6=0
=> x^2+3x+2x+6=0
=>x(x+3) +2(x+3)=0
=>(x+3)(x+2)=0
therefore,either x+3=0 or x+2=0
Hence x= -3 or -2
3.3x^2-10x-8=0
=>3x^2-12x+2x-8=0
=>3x(x-4)+2(x-4)=0
=>(x-4)(3x+2)=0
Hence either x-4=0 or 3x+2=0
Therefore,either x=4 or -2/3

Answer 2

1. take the sqrt of both sides and you get x=sqrt of 16=4,-4
2.(x+2)(x+3)=0 take the 2 and the 3 to the side with the 0
and subtract 2 from the first and 3 from the second and you get
-2,-3 as the answer
3.(3x+2)(x-4)=0 make 2 equations from this problem and you get
3x+2=0 and x-4=0 and subtract 2 on both sides on the first one and 4 on both sides to the second one and you have 3x=-2 and x=4 In the first equation you will divide both sides by 3 and the
final solution is x=-2/3, 4

Answer 3

x^2 = 16
x = 4 or -4

x^2+5x+6 = 0
(x+3)(x+2) = 0
x= -3 or x= -2

3x^2-10x-8 = 0
(3x+2)(x-4) = 0
x= -3/2 or x=4

Answer 4

x² - 5x - 6 ........minuend 3x²+ 6x - 8 ........subtrahend ----------------- -2x² -11x + 2 ........enormous difference tip: even as subtracting you should combine like words and continually change the signal of the subtrahend, then carry out the operation.