Factorize 2x⁴+x³-14x²-19x-6
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(x + 1)(x + 2)(2x + 1)(x - 3).

Step-by-step explanation:

From the leading and last coefficients, 2 and -6  the values +/- 1, +/-2 , +/- 3 could be roots of this equation ( well, up to 4 of them).

Try x = 1:

f(1) = 2(1)^4 + 1^3 - 14(1)^2 - 19(1) - 6 = -36 so its not 1.

f(-1) = 2 - 1 - 14 + 19 - 6 = 0 do x = -1 is a root.

Therefore (x + 1) is a factor  (By the factor theorem)

Try x = 2:

f(2) =  32 + 8 -56 - 38 - 6 = -6o so x =2 not a root.

f(-2) = 32 - 8 - 56 + 38 - 6 = 0 so x = -2 is a root and (x + 2) is a factor.

(x + 1)(x + 2) = x^2 + 3x + 2.

Long division:

                        2x^2 - 5x - 3   <Quotient

x^2 + 3x + 2 )     2x^4  +   x^3   -14x^2 -  19x  -  6

                          2x^4  + 6x^3 +  4x^2  

                                      -5x^3  - 18x^2 - 19x

                                      -5x^3  - 15x^2 - 10x

                                                      -3x^2 - 9x  - 6

                                                      -3x^2  - 9x - 6

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

So Finally,   the factors are (x + 1)(x + 2)(2x + 1)(x - 3).

no he doesnt because "-16' is negatiive

step-by-step explanation:

nope.   happens to me as well.

-aparri