solv it by another method without
without taking y = x square - 4x
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TheIncorporealKlaus:
Do by hit and trial method, make it a function, substitute the factors of 20 to see if (x-a) is a factor of f(x) and write the function as a quotient divisor product form, do this twice and you'll be left with a simple quadratic equation in terms of x, which you can resolve easily
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Given Equation is (x^2 - 4x)(x^2 - 4x - 1) - 20.
= > (x^2 * x^2 + x^2(-4x) + x^2(-1) - 4x * x^2 - 4x(-4x) - 4x(-1)) - 20
= > (x^4 - 4x^3 - x^2 - 4x^3 + 16x^2 + 4x )- 20
= > x^4 - 8x^3 + 15x^2 + 4x - 20.
Now,
= > x^4 - 9x^3 + x^3 + 24x^2 - 9x^2 -20x - 24x - 20
= > (x + 1)(x^3 - 9x^2 + 24x - 20)
= > (x + 1)(x^3 - 7x^2 - 2x^2 + 10x + 14x - 20)
= > (x + 1)(x - 2)(x^2 - 7x + 10))
= > (x + 1)(x - 2)(x^2 - 2x - 5x + 10))
= > (x + 1)(x - 2)(x[(x - 2) - 5(x - 2))
= > (x + 1)(x - 2)(x - 2)(x - 5)
= > (x + 1)(x - 2)^2(x - 5)
Hope this helps!
= > (x^2 * x^2 + x^2(-4x) + x^2(-1) - 4x * x^2 - 4x(-4x) - 4x(-1)) - 20
= > (x^4 - 4x^3 - x^2 - 4x^3 + 16x^2 + 4x )- 20
= > x^4 - 8x^3 + 15x^2 + 4x - 20.
Now,
= > x^4 - 9x^3 + x^3 + 24x^2 - 9x^2 -20x - 24x - 20
= > (x + 1)(x^3 - 9x^2 + 24x - 20)
= > (x + 1)(x^3 - 7x^2 - 2x^2 + 10x + 14x - 20)
= > (x + 1)(x - 2)(x^2 - 7x + 10))
= > (x + 1)(x - 2)(x^2 - 2x - 5x + 10))
= > (x + 1)(x - 2)(x[(x - 2) - 5(x - 2))
= > (x + 1)(x - 2)(x - 2)(x - 5)
= > (x + 1)(x - 2)^2(x - 5)
Hope this helps!
Done!
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