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Kay [80]
3 years ago
12

Find a polynomial of the specified degree that has the given zeros. Degree 4; zeros −1, 1, 4, 5 P(x) =

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
6 0

Answer:

The polynomial is;

x^4-9x^3+19x^2+9x-20

Step-by-step explanation:

Since it has a degree of 4, the highest power is 4

We have four zeros

So this means that each of the linear factors are;

We get this by equating each of the terms to x

(x-1)(x+1)(x-4)(x-5)

(x-1)(x+1) = x^2-1

So we have

(x-4)(x^2-1)

= x^3-x-4x^2 + 4

And lastly, we have

x(x^3-x-4x^2 + 4)-5(x^3-x-4x^2+4)

So we have;

x^4-x^2-4x^3+4x-5x^3+5x+20x^2-20

x^4-9x^3+19x^2+9x-20

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The perimeter of a rectangle is 72 cm. The length is 3 more than twice the width What
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<h3><u>Solution</u></h3>

<u>Given </u><u>:</u><u>-</u>

  • Perimeter of rectangle = 72 cm
  • The length is 3 more than twice the width.

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>:</u><u>-</u>

  • Dimensions of rectangle
<h3 /><h3><u>Explantion</u></h3>

<u>Using </u><u>Formula</u>

\boxed{\underline{\tt{\red{\:(perimeter_{rectang}\:=\:2\times (Length+Width)}}}}

<u>Let,</u>

  • Length of Rectangle = x cm
  • Breadth of Rectangle = y cm

<u>According</u><u> to</u><u> question</u><u>,</u>

==> perimeter of Rectangle = 72

==> 2(x+y) = 72

==> x + y = 72/2

==> x + y = 36_________________(1)

<u>Again,</u>

==> x = 2y + 3

==> x - 2y = 3__________________(2)

<u>Subtract</u><u> </u><u>equ(</u><u>1</u><u>)</u><u> </u><u>&</u><u> </u><u>equ(</u><u>2</u><u>)</u>

==> y + 2y = 36 - 3

==> 3y = 33

==> y = 33/3

==> y = 11

<u>keep </u><u>in </u><u>equ(</u><u>1</u><u>)</u>

==> x - 2×11 = 3

==> x = 3 + 22

==> x = 25

<h3><u>Hence</u></h3>

  • <u>Length</u><u> of</u><u> </u><u>Rectangle</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u> </u><u>cm</u>
  • <u>Width </u><u>of </u><u>Rectangle</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u> </u><u>cm</u>

<h3><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u></h3>

<h3 />

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