(x²-4x)(x²-4x-1)-20 is factorised as
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(x^2-4x)(x^2-4x-1)-20
x^2(x^2-4x-1)-4x(x^2-4x-1)-20
(x^4-4x^3-x^2) - (4x^3+16x^2+4x)-20
x^4-4x^3-x^2-4x^3-16x^2-4x-20=0
(x^4-8x^3-17x^2-4x)-20=0
X(x^3-8x^2-17x-4)-20=0
(x^3-8x^2-17x-4)-20=0/X
(x^3-8x^2-17x-4)-20=0
x^3-8x^2-17x-4-20=0
x^3-8x^2-17x-24=0
so prove that...
ax^3-8x^2-17x-24=0
this is your answer..
x^2(x^2-4x-1)-4x(x^2-4x-1)-20
(x^4-4x^3-x^2) - (4x^3+16x^2+4x)-20
x^4-4x^3-x^2-4x^3-16x^2-4x-20=0
(x^4-8x^3-17x^2-4x)-20=0
X(x^3-8x^2-17x-4)-20=0
(x^3-8x^2-17x-4)-20=0/X
(x^3-8x^2-17x-4)-20=0
x^3-8x^2-17x-4-20=0
x^3-8x^2-17x-24=0
so prove that...
ax^3-8x^2-17x-24=0
this is your answer..
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