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Aleksandr-060686 [28]
2 years ago
11

Question 20

Mathematics
1 answer:
Sveta_85 [38]2 years ago
8 0
I don’t know bro lol
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Consider the function f(x)=3x^3-5x^2-88x+60-5 is a root of f(x). Find all of the roots of f(x)
Tom [10]
If -5 is a root then f(x) is divisible by x + 5

The quotient from this division is 3x^2 -20x +12
Factoring this:-

= 3x^2 - 18x - 2x + 12
= 3x(x - 6) - 2(x - 6)
= (3x - 2)(x - 6)
So 2 more roots are 2/3 and 6
All the roots are -5 , 2/3 and 6.  Answer
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3 years ago
Seventy-two percent of the shareholders in a service corporation are women. If the corporation is
babunello [35]

Answer:

34416

Step-by-step explanation:

.72(percentage) × 47800(people)

34416

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2 years ago
Please help A.S.A.P. What possible transforms are shown below? (Choose at least one.)
n200080 [17]
3 reflection of y axis
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3 years ago
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If X = 2 centimeters, Y = 4 centimeters, and Z = 6 centimeters, what is the area of the object?
Tatiana [17]

Answer:

Step-by-step explanation:

the shape is a trpizoid with height y

A=(a+b)/2 *h (a=2x=4cm, h=y=4cm, b=2z=12 cm

A=(4+12)/2 * 4

Area=16/2 *4=8*4= 32 cm²

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3 years ago
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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.
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
1 year ago
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