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3 years ago

Solving Polynomials (PLZ HELP!! BRAINLIEST IF RIGHT)

Mathematics

2 answers:

givi [52]3 years ago

6 0

Answer:

$x=\frac{2}{5}$

Step-by-step explanation:

$5x^3-2x^2+5x-2=0\\x^2(5x-2)+(5x-2)=0\\(x^2+1)(5x-2)=0\\Hence,\\Either\ (x^2+1)\ should\ equal\ zero, or (5x-2)\ should\ be\ null.\\Lets\ try\ each\ on\ seperately:\\(x^2+1)=0\\x^2=-1\\x=\sqrt{-1}\\As\ \sqrt{-1}\ is\ not\ a\ real\ number,\ we\ can\ exclude\ it\ as\ a\ solution.\\Hence,\\(5x-2)=0\\5x=2\\x=\frac{2}{5} \\Thats\ your\ answer.$

Bas_tet [7]3 years ago

5 0

Answer:

x = $\frac{2}{5}$ , x = ± i

Step-by-step explanation:

Given

5x³ - 2x² + 5x - 2 = 0 ( factor the first/second and third/fourth terms )

x²(5x - 2) + 1(5x - 2) = 0 ← factor out (x - 2) from each term

(5x - 2)(x² + 1) = 0

Equate each factor to zero and solve for x

5x - 2 = 0 ⇒ 5x = 2 ⇒ x = $\frac{2}{5}$

x² + 1 = 0 ⇒ x² = - 1 ⇒ x = ± $\sqrt{-1}$ = ± i

The polynomial is of degree 3 and has 3 roots, 1 real and 2 complex, that is

x = $\frac{2}{5}$ ← real root

x = ± i ← complex roots

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