₦airaland Forum

Factorise the given expression [ see solution http://samimath.com/q-a-factorization-2/]

Do people still solve this things?

After equating d expression to zero:
3^2 = (x^2 - 3x - 1)^2
Removing squares:
3= (x^2 - 3x - 1)
After opening bracket;
3+1 = x^2 -3x
4 = x^2 - 3x
x^2 - 3x -4=0
Getting values for 3x dat is common to x^2 & -4
x^2 - 4x + x - 4 = 0
-:, x(x - 4) + 1(x-4) = 0
(x+1) = 0 & (x-4) = 0
Therefore:
x = -1
x = 4

the final answer is -8.

4kings:
After equating d expression to zero:
3^2 = (x^2 - 3x - 1)^2
Removing squares:
3= (x^2 - 3x - 1)
After opening bracket;
3+1 = x^2 -3x
4 = x^2 - 3x
x^2 - 3x -4=0
Getting values for 3x dat is common to x^2 & -4
x^2 - 4x + x - 4 = 0
-:, x(x - 4) + 1(x-4) = 0
(x+1) = 0 & (x-4) = 0
Therefore:
x = -1
x = 4

Damn! Did you see an equal to sign there that you are getting specific answers? Don't you know what factorisation is?

FarukuWadata12:
the final answer is -8.

Factorise is what the queation says. Damn!

-(x-1)(x+1)(x-2)(x-4)

Secondary school days were all fun for me

naturalwaves:
Damn! Did you see an equal to sign there that you are getting specific answers? Don't you know what factorisation is?

Hmm, then in that case I think the solution will be
3^2 - (x^2 - 3x - 1)^2

Going further will get u nowhere(though am not 100% sure)

4kings:
Hmm, then in that case I think the solution will be
3^2 - (x^2 - 3x - 1)^2

Going further will get u nowhere(though am not 100% sure)

Okay. Will get a pen once I settle down.

naturalwaves:
Okay. Will get a pen once I settle down.

I'm waiting

AnalyticEngine:
-(x-1)(x+1)(x-2)(x-4)

Please can u elucidate.

4kings:
Please can u elucidate.

9-(xsq-3x-1)sq = 3sq - (xsq -3x -1)sq
=(3+xsq-3x-1)(3-xsq+3x+1) ...differences of squares
= (2-3x+xsq)(4+3x-xsq) = -(2-3x+xsq)(xsq-3x-4) = -(2-3x+xsq)(x-4)(x+1) = -(x-2)(x-1)(x-4)(x+1) or (2-x)(x-1)(x-4)(x+1)

AnalyticEngine:
9-(xsq-3x-1)sq = 3sq - (xsq -3x -1)sq

=(3+xsq-3x-1)(3-xsq+3x+1) ...differences of squares

= (2-3x+xsq)(4+3x-xsq)
= -(2-3x+xsq)(xsq-3x-4)
= -(2-3x+xsq)(x-4)(x+1)
= -(x-2)(x-1)(x-4)(x+1) or (2-x)(x-1)(x-4)(x+1)

Dammmmmmmmmmm,
I can't believe I missed that(difference of square). Thanks.

it pains me when people call these stupid ass equations hard. What will you now do when you come across calculus or abstract algebra. Btw, i think the answer is (x-4) (x+1)

it's not hard, it's annoying

modified

9 - (x^2 - 3x - 1)^2


= 3^2 - (x^2 - 3x - 1)^2


difference of two squares

(3 - (x^2 - 3x - 1)) ( 3 + (x^2 - 3x - 1))

(3 - x^2 + 3x + 1)( 3 + x^2 - 3x - 1)

( - x^2 + 3x + 3 + 1)( x^2 - 3x + 3 - 1)

= ( - x^2 + 3x + 4)( x^2 - 3x + 2)

= ( - (x^2 - 3x - 4)) ( x^2 - 3x + 2)


= ( - ((x - 4)(x + 1)) ((x - 2)(x - 1))

it's actually

= - (x - 4)(x + 1)(x - 2)(x - 1)

AnalyticEngine:
9-(xsq-3x-1)sq = 3sq - (xsq -3x -1)sq

=(3+xsq-3x-1)(3-xsq+3x+1) ...differences of squares

= (2-3x+xsq)(4+3x-xsq)
= -(2-3x+xsq)(xsq-3x-4)
= -(2-3x+xsq)(x-4)(x+1)
= -(x-2)(x-1)(x-4)(x+1) or (2-x)(x-1)(x-4)(x+1)

damn difference of two squares...it's annoying