Mathematics, 18.10.2021 08:40 rach1314

Can anyone tell my how to find the zeros of the function: x^5-7x^4+19x^3-37x^2+60x-46. I am confused on how to factor this function.

Can anyone tell my how to find the zeros of the function: x^5-7x^4+19x^3-37x^2+60x-46. I am confuse

Answers: 3

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Answer from: robert7248

The zero is at x=3.497

Step-by-step explanation:

This polynomial can be simplified by factoring by grouping. The groups for this polynomial are:

$(x^{5}-7x^{4})+(19x^{3}-37x^{2})+(60x-46)=0$

For each group, factor out the common terms:

$(x^{5}-7x^{4})$ becomes $x^{4}(x-7)$

$(19x^{3}-37x^{2})$ becomes $x^{2}(19x-37)$

(60x-46) becomes 2(30x-23)

The original equation has factored into: $x^{4}\left(x-7\right)+x^{2}\left(19x-37\right)+2\left(30x-23\right)$

Since the factored equation does not have any common terms, it can't be factored anymore.

There are several ways to find the zeros of this function. The simplest way would be to graph the function and determine where it intersects the x-axis. Assuming you need to find the zeros without graphing the function, you can use synthetic division or approximate to find the zeros.

Using approximation and assuming the zeros are between 0 and 5:

$x^{4}\left(x-7\right)+x^{2}\left(19x-37\right)+2\left(30x-23\right)=0$

Testing x=0:

0(0-7)+0(0-37)+2(0-23)=2*-23=-46

Testing x=1:

1(1-7)+1(19*1-37)+2(30*1-23)=-6-18+14=-10

Testing x=2:

$2^{4}(2-7)+2^{2}(19*2-37)+2(30*2-23)=-80+4+74=-2$

Testing at x=3:

$3^{4}(3-7)+3^{2}(19*3-37)+2(30*3-23)=-324+180+134=-10$

There potentially is a zero between x=2 and x=3. For this function, there is not a zero there. This can be determined by testing more points between x=2 and x=3 or by graphing the function. The easiest extra points to test are x=1.9 and x=2.1. If both of these produce a value less than -2 then there is not a zero between x=2 and x=3 because x=2 would be part of the vertex of this graph segment.

Testing at x=4:

$4^{4}(4-7)+4^{2}(19*4-37)+2(30*4-23)=-768+624+194=54$

There is a potential zero between x=3 and x=4.

Testing at x=3.5:

$3.5^{4}(3.5-7)+3.5^{2}(19*3.5-37)+2(30*3.5-23)=-525.2188+361.375+164=0.1562$

This means that the function has a zero very close to 3.5. Additional points to test to approximate this zero are 3.4 and 3.6. You can continue to pick bracketing points to test until you are satisfied. When graphed, this zero is actually x=3.497. This is actually the only zero this function has when you look at its graph. There is no need to test any more of its points, and you can verify that this function is rapidly increasing once it passes through x=3.

Answer from: Quest

probability for j is 1/4

while

a is 1/6

Answer from: Quest

answer is

93.98 sq. in.

hope you got it chief

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Can anyone tell my how to find the zeros of the function: x^5-7x^4+19x^3-37x^2+60x-46. I am confused...

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