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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial