Factor f(x)=2x^4+19x ^3+37x^2-55x-75 and find the real roots.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find the zeros of f(x)=2x^4+19x^3+37x^2-55x-75

f(x)=2x^4+19x^3+37x^2-55x-75

=> 2x^4 + 2x^3 + 17x^3 + 17x^2 + 20x^2 + 20x - 75x - 75

=> 2x^3(x + 1) + 17x^2(x + 1) + 20x(x + 1) - 75(x + 1)

=> (x + 1)(2x^3 + 17x^2 + 20x - 75)

=> (x+ 1)(2x^3 + 10x^2 + 7x^2 + 35x - 15x - 75)

=> (x+ 1)(2x^2(x + 5) + 7x(x + 5) - 15(x + 5))

=> (x+ 1)(2x^2 + 7x - 15)(x + 5)

=> (x + 1)(x + 5)(2x^2 + 10x - 3x - 15)

=> (x + 1)(x + 5)(2x(x + 5) - 3(x + 5))

=> (x + 1)(x + 5)(2x - 3)(x + 5)

=> (x + 1)(x + 5)^2*(2x - 3)

Equating this to zero

(x + 1)(x + 5)^2*(2x - 3) = 0

x = -1, x = -5 and x = 3/2

The factorized form of f(x)=2x^4+19x^3+37x^2-55x-75 =  (x + 1)(x + 5)^2*(2x - 3) and the roots are x = -1, x = -5 and x = 3/2

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