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Nadya [2.5K]
3 years ago
14

Perform the indicated operation to write given polynomial in standard form: (x^3+2x^2 − 3x − 1) +(4 − x − x^3)

Mathematics
1 answer:
Rzqust [24]3 years ago
5 0

Answer:

\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-4x+3

Step-by-step explanation:

To write any polynomial in standard form, you look at the degree of each term. You then write each term in order of degree, from highest to lowest, left to write.

To add the polynomials you need to:

Remove parentheses

\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=x^3+2x^2-3x-1+4-x-x^3

Group like terms

\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=x^3-x^3+2x^2-3x-x-1+4

Add similar elements

\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-3x-x-1+4\\\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-4x-1+4\\\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-4x+3

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Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

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Let us try to find out \angle E and \angle DFE. After that we will compare them.

<u>Finding </u>\angle DFE<u>:</u>

Side EG is a straight line so \angle GFE = 180

\angle GFE is sum of internal \angle DFE and external \angle DFG

\angle GFE = 180 = \angle DFE  + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)

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<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external \angle DFG = \angle E + \angle EDF

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It can be clearly seen that:

\angle DFE = \angle E =90-\dfrac{y}{2}

The two angles of \triangle EDF are equal hence \triangle EDF is isosceles.

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