Question

solve the equation 2x^3-x^2-22x-24=0 given that two of its roots are in the ratio 3:4​

Answer 1

Step-by-step explanation:

2x³-x²-22x-24 = 0

(x+2) is one its factor by putting the value of x = -2

through hit and trial method you can find it to be the one of the factor.

Now divide the equation by (x+2) to get the other zeroes.

(x+2) | 2x³-x²-22x-24 | 2x²

2x³+4x²

(x+2) | 5x²-22x-24 | 2x²+5x

5x²+10x

(x+2) | -12x-24 | 2x²+5x-12

-12x-24

00 00

2x²+5x-12

2x²+8x-3x-12

2x(x+4) -3 (x+4)

(x+4) (2x-3)

x = -4

& x = 3/2

the roots of the equation are...

(-2, -4, 3/2)

Hope it will help you!

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