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allsm [11]
4 years ago
15

Divide: 8xy2 - 14y2 2y

Mathematics
2 answers:
yanalaym [24]4 years ago
8 0

Answer:

y(4x-7)  

Step-by-step explanation:

We have been given an expression 8xy^2-14y^2 and we are asked to divide our given expression by 2y.

Upon writing our given problem as a division problem we will get,

\frac{8xy^2-14y^2}{2y}

Upon factoring out the greatest factor of the terms from numerator we will get,

\frac{2y^2(4x-7)}{2y}  

y(4x-7)  

y(4x-7)    

Therefore, the quotient of our given problem is y(4x-7).

eimsori [14]4 years ago
3 0
<span>For this case we have the following expression:
</span> (8xy^2 - 14y^2)/2y&#10;
<span> Rewriting the expression we have:
</span> (4xy ^ 2 - 7y ^ 2) / y &#10;
<span> Then, for power properties we have:
</span> (4xy ^ {(2-1)} - 7y ^ {(2-1)}) &#10;
<span> Rewriting the expression we have:
</span> (4xy ^ {(1)} - 7y ^ {(1)}) &#10;
 4xy - 7y &#10;
<span> Answer:
 The equivalent expression is given by:
 4xy - 7y</span>
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The average price of a 30 second commercial for the 2002 Super Bowl was $1,900,000. What was the price per second?
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Find a factorization of x² + 2x³ + 7x² - 6x + 44, given that<br> −2+i√√7 and 1 - i√/3 are roots.
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A factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

<h3>What are the properties of roots of a polynomial?</h3>
  • The maximum number of roots of a polynomial of degree n is n.
  • For a polynomial with real coefficients, the roots can be real or complex.
  • The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if a+ib is a root, then a-ib is also a root.

If the roots of the polynomial p(x)=ax^4+bx^3+cx^2+dx+e are r_1,r_2,r_3,r_4, then it can be factorized as p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4).

Here, we are to find a factorization of p(x)=x^4+2x^3+7x^2-6x+44. Also, given that -2+i\sqrt{7} and 1-i\sqrt{3} are roots of the polynomial.

Since p(x)=x^4+2x^3+7x^2-6x+44 is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, -2-i\sqrt{7} and 1+i\sqrt{3} are also roots of the given polynomial.

Thus, all the four roots of the polynomial p(x)=x^4+2x^3+7x^2-6x+44, are: r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}.

So, the polynomial p(x)=x^4+2x^3+7x^2-6x+44 can be factorized as follows:

\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)

Therefore, a factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

To know more about factorization, refer: brainly.com/question/25829061

#SPJ9

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