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2 years ago

What difficulty did you encounter in determining polynomials?

Mathematics

1 answer:

liubo4ka [24]2 years ago

8 0

Answer:

I didn’t really encounter any difficulty in determining a polynomial. The issue was that numbers with negative exponents or those which had square roots in them aren’t referred to as polynomials.

The only minor issue was solving problems with numbers which had different exponents in them and needed some form of simplification.

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What are the zeros of the quadratic function f(x) = 6x^2 + 12x – 7?

strojnjashka [21]

Answer:

$x1=\frac{-2+\sqrt{26/3}}{2}$

$x2=\frac{-2-\sqrt{26/3} }{2}$

Step-by-step explanation:

To find the zeros of the quadratic function f(x)=6x^2 + 12x – 7 we need to factorize the polynomial.

To do so, we need to use the quadratic formula, which states that the solution to any equation of the form ax^2 + bx + c = 0 is:

$x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

So, the first thing we're going to do is divide the whole function by 6:

6x^2 + 12x – 7 = 0 -> x^2 + 2x - 7/6

This step is optional, but it makes things quite easier.

Then we using the quadratic formula, where:

a=1, b= 2, c = -7/6.

Then:

$x=\frac{-2±\sqrt{2^{2}-4(1)(-7/6)}}{2}$

$x=\frac{-2±\sqrt{4 +14/3}}{2}$

$x=\frac{-2±\sqrt{26/3}}{2}$

So the zeros are:

$x1=\frac{-2+\sqrt{26/3}}{2}$

$x2=\frac{-2-\sqrt{26/3}}{2}$

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1 year ago

What difficulty did you encounter in determining polynomials?​

What difficulty did you encounter in determining polynomials?