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Cerrena [4.2K]
3 years ago
9

How can you quickly determine the number of roots a polynomial will have by looking at the equation?

Mathematics
2 answers:
Goryan [66]3 years ago
7 0
First, you need to get delta which is b squared minus 4ac. If delta is higher than cero, the polynomial has two solutions, if it is less, the polynomial has no real solutions and if it is the same, it has one solution.
Although that is only for second grade ecuations
worty [1.4K]3 years ago
6 0

Answer:

One can determine the number of roots by seeing the degree of the  given polynomial.

Step-by-step explanation:

The number of roots can be determine by just seeing the highest power of the given equation.

Example,

1) for the equation  x-4=0

here, the highest power the equation is one. So, it will have one root.

Lets check it by simplify  x-4=0 \Rightarrow x=4

Hence, the equation has only one root namely x = 4.

2) Consider a quadratic equation,

10t^2-t-3=0

here, the highest power the equation is two.So, it will have two roots.

Lets check it by simplify using middle term splitting method,

10t^2+5t-6t-3=0

5t(2t+1)-3(2t+1)=0

(2t+1)(5t-3)=0

\Rightarrow t=\frac{-1}{2} or \Rightarrow t=\frac{3}{5}

thus, the equation has two roots.

Hence, one can determine the number of roots by seeing the degree of the  given polynomial.

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PLEASE HELP
nydimaria [60]

Answer:

y = -2(x+2)(x-1)/((x+3)(x-6)) = (-2x^2 -2x +4)/(x^2 -3x -18)

Step-by-step explanation:

A polynomial function will have a zero at x=a if it has a factor of (x-a). For the rational function to have zeros at x=-2 and x=1, the numerator factors must include (x+2) and (x-1).

For the function to have vertical asymptotes at x=-3 and x=6, the denominator of the rational function must have zeros there. That is, the denominator must have factors (x+3) and (x-6). Then the function with the required zeros and vertical asymtotes must look like ...

f(x) = (x+2)(x-1)/((x+3)(x-6))

This function will have a horizontal asymptote at x=1 because the numerator and denominator degrees are the same. In order for the horizontal asymptote to be -2, we must multiply this function by -2.

The rational function may be ...

y = -2(x +2)(x -1)/((x +3)(x -6))

If you want the factors multiplied out, this becomes

y = (-2x^2 -2x +4)/(x^2 -3x -18)

6 0
3 years ago
What is one factor of 5x^2+13x-6 with work please.
iren [92.7K]

Answer:

Factor: 5x^2+13x−6

5x^2+13x−6

<u>=(5x−2)(x+3)</u>

<h3><em><u>Brainliest please?</u></em></h3>
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3 years ago
NEED THE ANSWER ASAP THANKS!!!!!!
iVinArrow [24]

Answer:

its a

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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