Determine the sign of polynomial 2x^2-5x-3 for all real values of x.?

Answer 1

2x² - 5x - 3 = 0

2x² - 6x + 1x - 3 = 0

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

(2x + 1)(x - 3)

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Roots

2x + 1 = 0

2x + 1 - 1 = 0 - 1

2x = - 1

2x/2 = - 1/2. . <=. .not used

x = - 1/2

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X - 3 = 0

x - 3 + 3 + 0 + 3

x = 3

The answer is x = 3

Insert the x value into the euation

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2x² - 5x - 3 = 0

2(3)² - 5(3) - 3 = 0

2(9) - 15 - 3 = 0

18 - 18 = 0

0 = 0

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Answer 2

P(x)=2x^2-5x-3
The polynomial in U shape because the coefficient of x^2 is positive.
Let's find the points in which P(x) is 0.

2x^2-5x-3=0
x1=-0,5
x2=3
Between x=-0,5 and x=3, the sign of the polynomial is negative.
When x=-0,5, the polynomial is 0.
When x=3, the polynomial is 0.

When x<-0,5, the sign of the polynomial is positive.
When x>3, the sign of the polynomial is positive.

Answer 3

OK, set the polynomial equal to zero, solve for x

(2x²-5x-3)=0

(x-3)(2x+1)=0

x=3, x=-1/2

At those points, x is zero. Between them, the curve is either positive or negative' outside of them, it's the opposite.
0 is between those points. At x=0, f(x) is -3.

So, curve is positive (-∞, -1/2) and (3,∞), negative on (-1/2,3) and zero as listed above.

Answer 4

y = 0 <=> x = 3 or x = -1/2
with every real values of x within (-1/2,3) the sign of polynomial will be negative, and will be positive with the rest

Answer 5

sign of the given polynomial is positive for all real values of x

Answer 6

(2x+1)(x-3) zero at -1/2 & at +3

x<-1/2, +
x>3, +
elsewere -

Answer 7

I'd just go with pi, thats usually a pretty good guess for any math question

Answer 8

2x^2-5x-3
2x^2-6x+1x-3
(2x^2-6x)+(1x-3)
2x(x-3)+(x-3)
(2x+1)(x-3)

x= 3 and -1/2