For the polynomial below,what is the power of the term with the coefficient of 1? X^3+1/3x^4+6x+5

Mathematics

2 answers:

julia-pushkina [17]3 years ago

8 0

Answer:

It's 3

Step-by-step explanation:

valentina_108 [34]3 years ago

5 0

Original Polynomial
${x}^{3} + \frac{1}{ {3x}^{4} } + 6x + 5$
Coefficient Concept Expansion
$1 \times {x}^{3} = {x}^{3}$
The Power of the term with 1 as the coefficient is 3.

A coefficient is a number multiplied by some variable. For instance, 4x has a coefficient of 4 or 56x has a coefficient of 56.

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Which are the solutions of x^2 = –11x + 4?

strojnjashka [21]

This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

$x^2 + 11x - 4 = 0$

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}$

where $x_{1,2}$ is a compact way to indicate both solutions $x_1$ and $x_2$ , while $a,b,c$ are the coefficients of the quadratic equation, i.e. we consider the polynomial $ax^2+bx+c$ .

So, in your case, we have $a=1,\ \ b=11,\ \ c=-4$

Plug those values into the formula to get

$x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}$

So, the two solutions are

$x_1 = \frac{-11+\sqrt{137}}{2}$

$x_2 = \frac{-11-\sqrt{137}}{2}$