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WITCHER [35]
4 years ago
13

Subtract polynomials

Mathematics
1 answer:
ziro4ka [17]4 years ago
6 0

Answer: For this, you need to equal the equations to each other, then combine like terms and subtract from both sides :).

Step-by-step explanation:

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What is the answer for -1/9
chubhunter [2.5K]
I hope this helps you

4 0
4 years ago
Convert <br>431 base 10 divided by 112base 10 Then u should convert the final answer to base 5​
Rainbow [258]

Answer:

32115

Step-by-step explanation:

Step1

Converting 431 to base 10.

4*102=400

3*101=30

1*100=1

Adding all to get Ans=43110

Step2 converting 43110 to 5

The equation calculation formula for 43110 number to 5 is like this below.

5|431

5|86|1

5|17|1

5|3|2

5|3|3

Ans:32115

5 0
3 years ago
A television normally cost $1420 due to a store sale the television is now $1022.40 what was the percentage taken off the televi
Ymorist [56]

Answer:

28% was taken off the price

Step-by-step explanation:


4 0
3 years ago
4 1/6 - 3 3/5<br> Show your work pls
ki77a [65]

Answer:

-78/30 or -2 3/5

Step-by-step explanation:

4 1/6 - 3 3/5

25/6 - 18/5

30-108\30

= -78/30 or -2 3/5

8 0
3 years ago
Read 2 more answers
Find a factorization of x² + 2x³ + 7x² - 6x + 44, given that<br> −2+i√√7 and 1 - i√/3 are roots.
Levart [38]

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

3 0
2 years ago
Read 2 more answers
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