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Factorize 2x⁴+x³-14x²-19x-6
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1 Answer

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Answer:

(x + 1)(x + 2)(2x + 1)(x - 3).

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).

User Ken Cochrane
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