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4x**2-3x-5

quadratic = (-b+sqrt(b**2-4ac))/2a

quadratic = (-b-sqrt(b**2-4ac))/2a

b**2-4ac>0 --> 2 roots

b**2-4ac=0 --> 1 root

b**2-4ac<0 --> 0 roots

(-3)**2-4(4*-5)

9+80=89

2 roots = (6+89**0.5)/8, (6-89**0.5)/8

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Q: How many roots does the polynomial have 4x2 - 3x - 5?
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Find the roots of the polynomial given 4x2 - 3x - 5?

No integer roots. Quadratic formula gives 1.55 and -0.81 to the nearest hundredth.


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4x2 + 3x - 6 is a second degree polynomial. Since the polynomial function f(x) = 4x2 + 3x - 6 has 2 zeros, it has 2 linear factors. Since we cannot factor the given polynomial, let's find the two roots of the equation 4x2 + 3x - 6 = 0, which are the zeros of the function. 4x2 + 3x - 6 = 0 x2 + (3/4)x = 6/4 x2 + (3/4)x + (3/8)2 = 6/4 + 9/64 (x + 3/8)2 = 105/64 x + 3/8 = &plusmn; &radic;(105/64) x = (-3 &plusmn; &radic;105)/8 x = -(3 - &radic;105)/8 or x = -(3 + &radic;105)/8 Thus, the linear factorization of f(x) = 4[x + (3 - &radic;105)/8][x + (3 + &radic;105)/8].


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What are the roots of the polynomial x2 plus 3x plus 5?

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