Prove that the sum of the angles of a quadrilateral is equal to four right angles
Answers
Answer:
We solve the question through the method of long division. When we divide a number by another number, we try to eliminate the largest number. This is common in polynomial division, as the target is to eliminate the highest degree term.
①P(x)\text{ by }G(x)P(x) by G(x)
\iff x^{3}-3x^{2}+5x-3=(x^{2}-2)(x-3)+7x+9⟺x
3
−3x
2
+5x−3=(x
2
−2)(x−3)+7x+9
②P(x)\text{ by }G(x)P(x) by G(x)
\iff x^{4}-3x^{2}+4x+5=(x^{2}-x+1)(x^{2}+x-3)+8⟺x
4
−3x
2
+4x+5=(x
2
−x+1)(x
2
+x−3)+8
③P(x)\text{ by }G(x)P(x) by G(x)
\iff x^{4}-5x+6=(-x^{2}+2)(-x^{2}-2)-5x+10⟺x
4
−5x+6=(−x
2
+2)(−x
2
−2)−5x+10
If we list them as quotients and remainders:-
①Quotient is x-3x−3 . Remainder is 7x+97x+9 .
②Quotient is x^{2}+x-3x
2
+x−3 . Remainder is 88 .
③Quotient is -x^{2}-2−x
2
−2 . Remainder is -5x+10−5x+10 .
Here, I included the process of your each polynomial division. The method is using only coefficients in polynomial division. It is simpler and takes less time. :)
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More information
Suppose we have divided the polynomial. If AA and BB are the dividend and divisor, QQ and RR are the quotient and remainder, we would get the following equation:-
\large\boxed{A=BQ+R}
A=BQ+R
Such type of equation is called the identity. The left and right hand side are always equal. This fact is used in dealing with polynomials.
Pls mark my answer as brainliest
