2 x 4 + 5 x 3 + 38 x 2 + 125 x − 300 = 0 2 x 4 + 5 x 3 − 12 x 2 + 50 x 2 + 125 x − 300 = 0 x 2 ( 2 x 2 + 5 x − 12 ) + 25 ( 2 x 2 + 5 x − 12 ) + 0 ( 2 x 2 + 5 x − 12 ) ( x 2 + 25 ) = 0 ⇕ 1 o 2 x 2 + 5 x − 12 = 0 ∨ 2 o x 2 + 25 = 0 1 o Δ = 5 2 − 4 ⋅ 2 ⋅ ( − 12 ) = 25 + 96 = 121 Δ = 121 = 11 x 1 = 2 ⋅ 2 − 5 − 11 = 4 − 16 = − 4 ; x 2 = 2 ⋅ 2 − 5 + 11 = 4 6 = 2 3 2 o x 2 = − 25 − f a l se A n s w er : x = − 4 or x = 2 3
The solutions to the equation 2 x 4 + 5 x 3 + 38 x 2 + 125 x − 300 = 0 are found using numerical methods, yielding the roots x = − 4 and x = 2 3 . These points indicate the values for which the polynomial equals zero. Further techniques regarding synthetic division and the Rational Root Theorem were suggested for exploration.
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