Math, asked by pardeepnarwal9, 11 months ago

one of the factor of 2x^4-5x^2+5x-2​

Answers

Answered by harshbhagat22
0

Step-by-step explanation:

Four solutions were found :

x =√-0.186 = 0.0 - 0.43144 i

x =√-0.186 = 0.0 + 0.43144 i

x =√ 2.686 = -1.63894

x =√ 2.686 = 1.63894

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((2 • (x4)) - 5x2) - 1 = 0

Step 2 :

Equation at the end of step 2 :

(2x4 - 5x2) - 1 = 0

Step 3 :

Trying to factor by splitting the middle term

3.1 Factoring 2x4-5x2-1

The first term is, 2x4 its coefficient is 2 .

The middle term is, -5x2 its coefficient is -5 .

The last term, "the constant", is -1

Step-1 : Multiply the coefficient of the first term by the constant 2 • -1 = -2

Step-2 : Find two factors of -2 whose sum equals the coefficient of the middle term, which is -5 .

-2 + 1 = -1

-1 + 2 = 1

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step 3 :

2x4 - 5x2 - 1 = 0

Step 4 :

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

4.1 Solve 2x4-5x2-1 = 0

This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using w , such that w = x2 transforms the equation into :

2w2-5w-1 = 0

Solving this new equation using the quadratic formula we get two real solutions :

2.6861 or -0.1861

Now that we know the value(s) of w , we can calculate x since x is √ w

Doing just this we discover that the solutions of

2x4-5x2-1 = 0

are either :

x =√ 2.686 = 1.63894 or :

x =√ 2.686 = -1.63894 or :

x =√-0.186 = 0.0 + 0.43144 i or :

x =√-0.186 = 0.0 - 0.43144 i

Four solutions were found :

x =√-0.186 = 0.0 - 0.43144 i

x =√-0.186 = 0.0 + 0.43144 i

x =√ 2.686 = -1.63894

x =√ 2.686 = 1.63894

Answered by shikha6276
0

Answer:

x = 1  \\ p(x) = 2 \times (1 )^{4}  - 5 \times (1 )^{2}  + 5 \times( 1) - 2 \\  \:  \:  \:  = 2 \times 1 - 5 \times 1 + 5 - 2 \\  \:  \:  \:   = 2 - 5 + 5 - 2 \\  \:  \:  \:  = 0 \\ x = 1 \\ x + 1 = 0 \\ x + 1is \:  \:a \: factor \: of \: p(x) \\ x + 1

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