# `f(x) = x^5 - 2x^4 + 8x^2 - 13x + 6` Use the Rational Root Theorem to detrmine all the zeros of the function. a)  (x-1)^2(x+2)(x -1 - i sqrt2)(x -1 +i sqrt2) b)  (x-1)^2(x+2)(x+1 - i sqrt2)(x...

`f(x) = x^5 - 2x^4 + 8x^2 - 13x + 6`

Use the Rational Root Theorem to detrmine all the zeros of the function.

a)  (x-1)^2(x+2)(x -1 - i sqrt2)(x -1 +i sqrt2)

b)  (x-1)^2(x+2)(x+1 - i sqrt2)(x + 1 + i sqrt2)

c)  (x+1)^2(x-2)(x - 1 - i sqrt2)(x - 1 + i sqrt2)

d)  (x+1)^2(x-2)(x + 1 - i sqrt2)(x + 1 + i sqrt2)

e)  (x-1)^2(x+2)(x-3)(x+1)

f)  (x-1)^3(x+2)(x-3)

g)  (x-1)(x+2)(x+1)^2(x-3)

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`f(x)=x^5 - 2x^4 + 8x^2 - 13x + 6`

In this function, the coefficient of leading term x^5 is 1 and the constant is 6. So, the possible roots are the factors of the quotient of `+-6/1` . These are `+-1` , `+-2` , `+-3` , and `+-6` .

To determine which of them are the roots, divide the the polynomial by the possible roots. Use synthetic division.

Let's start x=1.

`1` `|` `1`    `-2`      `0`        `8`     `-13`     `6`
`1`     `-1`      `-1`          ` 7`      `6`
________________________________
`1`     `-1`      `-1`        ` 7`      `-6`     `0`

Since the last number is zero, then x=1 is a root of the f(x). This means that f(x) can be factored as:

`f(x)=(x-1)(x^4 -x^3-x^2+7x-6)`

To factor x^4-x^3-x^2-7x-6 further, divide it by one of the possible roots of f(x). Use synthetic division again.

Let's try x=1 again.

`1` `|`   `1`     `-1`      `-1`         `7`     `-6`

`1`        `0`      `-1`         `6`

____________________________

`1`       `0`      `-1`         `6`        `0`

Hence, x=1 is a root of f(x) again. So, f(x) can be factored as:

`f(x)=(x-1)^2(x^3-x+6)`

To factor x^3-x+6 further, divide it by the possible roots of f(x).

Let's try x=-2.

`-2` `|`  `1`       `0`      `-1`         `6`

`-2`        `4`      `-6`

______________________

`1`    `-2`        `3`         `0`

So, x=-2 is a root f(x). And its factor would be:

`f(x)=(x+1)^2(x-2)(x^2-2x+3)`

The last factor x^2-2x+3 is a quadratic expression. To factor this further, apply quadratic formula.

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(-2)+-sqrt((-2)^2-4*1*3))/(2*1)=(2+-sqrt(-8))/2=(2+-2isqrt2)/2`

`x=1+-isqrt2`

So, the other roots are x=1+isqrt2 and x=1-isqrt2.

Therefore, the factor of the given polynomial is `f(x)=(x-1)^2(x+2)(x-1-isqrt2)(x-1+isqrt2)` .

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