Solve:-
(x2-4x)(x2-4x-1)-20
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Answered by
0
HEY!!
(x2 4x) (x2 4x 1) 20
Putx2 4x=t
⇒ (x2 4x) (x2 4x 1) 20 =t(t 1) 20 =t2t 20 =t2 5t+ 4t 20
=t(t 5) + 4 (t 5) = (t+ 4) (t 5)
= (x2 4x+ 4) (x2 4x 5)
= (x2 2x 2x+ 4) (x2 5x+x 5)
= [x(x 2) 2 (x 2)] [x(x 5) + 1 (x 5)]
= (x 2) (x 2) (x+ 1) (x 5)
(x2 4x) (x2 4x 1) 20
Putx2 4x=t
⇒ (x2 4x) (x2 4x 1) 20 =t(t 1) 20 =t2t 20 =t2 5t+ 4t 20
=t(t 5) + 4 (t 5) = (t+ 4) (t 5)
= (x2 4x+ 4) (x2 4x 5)
= (x2 2x 2x+ 4) (x2 5x+x 5)
= [x(x 2) 2 (x 2)] [x(x 5) + 1 (x 5)]
= (x 2) (x 2) (x+ 1) (x 5)
Answered by
0
hy
here is your answer
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Put x2 – 4x = t
⇒ (x2 – 4x) (x2 – 4x – 1) –20 = t (t – 1) – 20 = t2 – t – 20 = t2 – 5t + 4t – 20
= t (t – 5) + 4 (t – 5) = (t + 4) (t – 5)
= (x2 – 4x + 4) (x2 – 4x – 5)
= (x2 – 2x – 2x + 4) (x2 – 5x + x – 5)
= [x ( x – 2) – 2 (x – 2)] [x ( x – 5) + 1 (x – 5)]
= (x – 2) (x – 2) (x + 1) (x – 5)
here is your answer
==================
Put x2 – 4x = t
⇒ (x2 – 4x) (x2 – 4x – 1) –20 = t (t – 1) – 20 = t2 – t – 20 = t2 – 5t + 4t – 20
= t (t – 5) + 4 (t – 5) = (t + 4) (t – 5)
= (x2 – 4x + 4) (x2 – 4x – 5)
= (x2 – 2x – 2x + 4) (x2 – 5x + x – 5)
= [x ( x – 2) – 2 (x – 2)] [x ( x – 5) + 1 (x – 5)]
= (x – 2) (x – 2) (x + 1) (x – 5)
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