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2 years ago

P+q=1 and pq=-20. What are the values of p and q?

Mathematics

1 answer:

andrey2020 [161]2 years ago

6 0

Answer:

Step-by-step explanation:

5, -4

5-4=1

5(-4)=-20

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Find all the zeros of the polynomial x^4+x^3-34x^2-4x+120 if two of its zeros are 2,-2

miskamm [114]

Answer:

The zeroes are 2, -2, -6, 5.

Step-by-step explanation:

If 2 of the zeroes are 2 and -2 then (x - 2)(x + 2) is factor of the polynomial.

We can now do long division.

Note (x - 2)(x + 2) = x^2 - 4, so we have:

x^2 - 4 ( x^4 + x^3 - 34x^2 - 4x + 120 ( x^2 + x - 30 <--- Quotient

x^4 - 4x^4

x^3 - 30x^2 - 4x

x^3 -4x

-30x^2 + 120

x^2 + x - 30 = 0

(x + 6)(x - 5) = 0

x = -6, 5.

7 0

1 year ago

Read 2 more answers

I do not know how to solve this problem or know what to do with the exponents

horsena [70]

K, remember
(ab)/(cd)=(a/c)(b/d) or whatever
also
$(ab)^c=(a^c)(b^c)$
and
$x^{-m}= \frac{1}{x^m}$
and
$x^ \frac{m}{n}= \sqrt[n]{x^m}$
and
$( \frac{x}{y} )^m= \frac{x^m}{y^m}$
and
$(x^m)^n=x^{mn}$
and
(a/b)/(c/d)=(a/b)(d/c)=(ad)/(bc)

so
$( \frac{-7x^ \frac{3}{2} }{5y^4} )^{-2}$ =
$( \frac{-7}{5} )^{-2}( \frac{x^ \frac{3}{2} }{y^4} )^{-2}$ =
$( \frac{(-7)^{-2}}{5^{-2}} )( \frac{(x^ \frac{3}{2})^{-2} }{(y^4)^{-2}} )$ =
$( \frac{ \frac{1}{(-7)^2} }{ \frac{1}{5^2} } )( \frac{x^ \frac{-6}{2} }{y^{-8}} )$ =
$( \frac{ \frac{1}{49} }{ \frac{1}{25} } )( \frac{(x^{-3}) }{\frac{1}{y^8}} )$
$( \frac{ \frac{1}{49} }{ \frac{1}{25} } )( \frac{\frac{1}{x^3} }{\frac{1}{y^8}} )$ =
$(\frac{25}{49} )( \frac{y^8}{x^3}$ =
$\frac{25y^8}{49x^3}$

4 0

3 years ago

Write each expression in exponential form and find its value 1/6*1/6*1/6*1/6

olga55 [171]

Answer:

1/6^4.

Step-by-step explanation:

Multiply denominators.

6*6*6*6 = 1,296

Your answers are 1/6^4 and 1/1296.