# factorise-x4-(x-z)4

x4-(x-z)4

(x2)2-{[(x-z)2]2}

or, x2-(x-z)2

or, [x-(x-z)][x+(x-z)]

or, (x-x+z)(x+x-z)

or, z(2x-z)

• -5

how can (x2)2-{[(x-z)2]2} be equal to x2-(x-z)2 ????

• 2

i know right.. i gave the answer wrongly... sorry..

• -2

(x 2 ) 2 -[(x-z) 2 ] 2  can be  x 2 -(x-z)

• -2

x 4 – (xz)4

= (x 2)2 – [(xz)2]2

The given equation is of the form, a 2b 2 = (a + b) (a – b),

where a = x 2 and b = (xz)2

∴ (x 2)2 – [(xz)2]2 = [x 2 + (xz)2] [x 2 – (xz)2]

= [x 2 + x 2 + z 2 – 2xz] [x 2x 2z 2 + 2xz]

= [2x 2 + z 2 – 2xz] [2xz 2]

= [2x 2 – 2xz + z2] × z[2xz]

Hence, x 4 – (x – z)4 = z (2xz )(2x 2 – 2xz + z 2)

• 13
The expression x 4 – (xz)4 can be factorised as,
x 4 – (xz)4
= (x 2)2 – {(xz)2}2
= [x 2 – (xz)2] [x 2 + (xz)2] [a 2b 2 = (ab) (a + b)]
= [x – (xz)] [x + (xz)] [x 2 + x 2 + z 2 – 2xz]
[a 2b 2 = (ab) (a + b), (ab)2 = a 2 + b 2 –2ab]
= [xx + z)] [2xz] [x 2 + x 2 + z 2 – 2xz]
= z (2xz) [2x 2 + z 2 – 2xz]
• 19 • 3
find the factors and multiples of 5 • 1
Huttyg
• 1
• 1 • 0
tu pa
• 0 • -1

x4-(x-z)4

(x2)2-{[(x-z)2]2}

or, x2-(x-z)2

or, [x-(x-z)][x+(x-z)]

or, (x-x+z)(x+x-z)

or, z(2x-z)

• -1
• 0
Forget means what
• 1
z[2x-z]
• 0
Nahi
• 0
Na karu
• 0 • 0 • 0
Express as directed
• 0 • 0 • 0 • 0 • 0 • 0
Ex: ax?+?by)2?+ (bx?+?ay)2
= [(ax)2?+ 2(ax)(by) + (by)2] + [(bx)2?+ 2(bx)(ay) + (ay)2]
= (a2x2?+ 2abxy?+?b2y2) + (b2x2?+ 2abxy?+?a2y2)ax?+?by)2?+ (bx?+?ay)2
= [(ax)2?+ 2(ax)(by) + (by)2] + [(bx)2?+ 2(bx)(ay) + (ay)2]
= (a2x2?+ 2abxy?+?b2y2) + (b2x2?+ 2abxy?+?a2y2)
=?a2x2?+?b2x2?+?a2y2?+?b2y2?+ 4abxy
=?x2(a2?+?b2) +?y2(a2?+?b2) + 4abxy
= (x2?+?y2) (a2?+?b2) + 4abxy
=?a2x2?+?b2x2?+?a2y2?+?b2y2?+ 4abxy

=?x2 un x value of 4(a2?+?b2) +?y2(a2?+?b3) + 4abxy

= (x2?+?y2) (a2?+?b2) + 4abxy

?
The factorise of x4 in every examples

:4abxy the factorise =is equal 4in every is called closure property
• 0 • 0 • 0
X4-4x-4z=4z
• 0
x?-(x-Z)?
(X?)? - {(X-Z)?}? [(a?-b?) = (a+b) (a-b)
{ x? + (X-Z)}{X-(X-Z) }
{x? +x-Z} {X-X+Z}
{x?+x-Z} {Z}
Z{x?+X-Z}
• 0 