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2 years ago

Find a polynomial equation that has zeros at x = -4, x = 0, and a double root at x = 7.

Mathematics

1 answer:

Ksju [112]2 years ago

7 0

Answer:

Step-by-step explanation:

Given a polynomial f(x) with roots say x = a and x = b, then the factors are

(x - a) and (x - b) and f(x) is the product of the roots, that is

f(x) = (x - a)(x - b)

Here the roots are x = - 4, x = 0 and double root at x = 7

Thus factors are (x - (- 4)), (x - 0), that is (x + 4) and x

Given a double root at x = 7 the factor is (x - 7)², thus polynomial is

y = x(x + 4)(x - 7)² → D

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Simplify expression x8y-26/x14y-5 × x-39y-21

koban [17]

The simplified form of the expression is $x^{-45} y^{-42}$

<h3>Simplifying an expression</h3>

From the question, we are to simplify the given expression

The given expression is

$\frac{x^{8}y^{-26} }{x^{14}y^{-5} } \times x^{-39} y^{-21}$

The expression can be simplified as shown below

$\frac{x^{8}y^{-26} }{x^{14}y^{-5} } \times x^{-39} y^{-21}$

$\frac{x^{8} \times x^{-39} \times y^{-26} \times y^{-21} }{x^{14} \times y^{-5} }$

$\frac{x^{8+-39} \times y^{-26+-21} }{x^{14} \times y^{-5} }$

$\frac{x^{8-39} \times y^{-26-21} }{x^{14} \times y^{-5} }$

$\frac{x^{-31} \times y^{-47} }{x^{14} \times y^{-5} }$

Then,

$\frac{x^{-31} }{x^{14} } \times \frac{ y^{-47} }{ y^{-5} }$

$x^{-31-14} \times y^{-47--5}$

$x^{-31-14} \times y^{-47+5}$

$x^{-45} \times y^{-42}$

$x^{-45} y^{-42}$

Hence, the simplified form of the expression is $x^{-45} y^{-42}$

Learn more on Simplifying an expression here: brainly.com/question/2320607

#SPJ1

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1 year ago

How would u solve for this

FromTheMoon [43]

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artcher [175]

Answer:

Step-by-step explanation

Take the square root of both sides.

Add both sides by 5

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Natali5045456 [20]

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I hope this helps!

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