Verified answer

You definitely have to have (6 + 5i) as one of the zeros. (All complex roots come in conjugate pairs.)

And if the coefficients are rational, you also need

(-1 - sqrt 7) as a zero in order to cancel out the sqrt 7 in the coefficients.

The roots are:

6 - 5i, 6 + 5i (both complex)

-1 + √ 7, -1 - √ 7 (both real, but irrational)

EXPLANATION.

Well,

the fundamental theorem of algebra says that if you have a polynomial of degree n with rational coefficients, then this polynomial will have exactly n zeros (though some of them can be repeated).

These zeros can be real or complex, and if there are any complex zeros, they come in conjugate pairs and therefore there is always an even number (can also be 0) of complex zeros.

Your poly of deg 4 will have 4 zeros.

One of them is 6 - 5i. It is complex and there will necessarily be another complex zero, namely 6 + 5i.

Therefore you can write your polynomial

f(x) = ax^4 + bx³ + cx² + dx + e as

f(x)= a·(x - [6-5i])·(x - [6+5i])·(x - [-1 + √ 7])(x - [4th zero])

{If you can't see why we can do so, contact me for further explanation}

Multiply (x - [6-5i])·(x - [6+5i]) :

(x - [6-5i])·(x - [6+5i]) =

x² - [6-5i]·x - [6+5i]·x - [6-5i]·[6+5i] =

x² - 12x - 36 + (5i)² = {{ i² = -1}}

x² - 12x - 61

Now let's think about the other two factors in the polynomial. You know the third zero is -1 + √ 7. It is irrational (√ 7 is irrational, i.e. cannot be expressed as a fraction). But your polynomial has RATIONAL coefficients.

Therefore the 4th zero has to be such that (x - [-1 + √ 7])(x - [4th zero]) will have rational coefficients.

There is only 1 way how to effectively choose the 4th zero.

See your complex zero and its conjugate, when multiplied (above), the complex i has disappeared altogether?

If we choose the forth zero to be -1 - √ 7, then in

(x - [-1 + √ 7])(x - [-1 - √ 7]) the square root will disappear altogether.

(x - [-1 + √ 7])(x - [-1 - √ 7]) =

x² - [-1 + √ 7]·x - [-1 - √ 7]·x - [-1 + √ 7]·[-1 - √ 7] =

x² + 2x - 1 + 7 =

x² + 2x + 6

So your polynomial is

f(x)= a·(x - [6-5i])·(x - [6+5i])·(x - [-1 + √ 7])(x - [-1 - √ 7])

f(x) = a·(x² - 12x - 61)·(x² + 2x + 6 )

f(x) = a·(x^4 - 10x³ - 79x² - 194x - 366)

Where a can be any rational number.

Hope this helps!

a million. Use guy made branch to coach that x= -3 is a root of the equation x^3 - 3x^2 -13x + 15 = 0. Then use the effect to factor the polynomial thoroughly to this form: (x+A)(x+B)(x+C) and supply the fee of A, B, and C -3 | ....a million ......-3...... -13....... 15 ... ......... ....-3.... ....18.......-15 --------------------------------------... ... .....a million.......-6......... 5.........0 <==the rest = 0, so x + 3 is an element quotient = x^2 - 6x + 5 = (x - 5)(x - a million) so totally factored: (x + 3)(x - 5)(x - a million) A = 3, B = -5, C = -a million (or any mixture thereof) notice: we commonly write aspects as (x - A)(x - B)(x - C), the place A,B, and C are the zeros are roots 2.discover all rational zeros of the polynomial x^3 - 3x^2 -13x +15. record each rational 0 the comparable sort of circumstances as its multiplicity. as an occasion, if a rational 0 has multiplicity one, record it as quickly as; if it has multiplicity 2, record it two times. If there are no longer any rational zeros, enter None . watching the coefficients, and employing the factor theorem (if f(c) = 0, then (x - c) is an element and x = c is a root), f(a million) = 0, so (x - a million) is an element a million^3 - 3(a million^2) - 13(a million) + 15 = 0 now use guy made or long branch to discover the quotient a million.|....a million... -3..... -13.... ..15 ..... ...........a million........-2.......-15 -------- ---------------------------- --- ....... a million..... -2...... -15..... ..0 <== the rest = 0, so x - a million is an element x^2 - 2x - 15 = quotient = (x - 5)(x - 3) so totally factored: (x - a million)(x - 5)(x - 3) rational zeros (a number of those zeros are rational: a million,3,5 3. For the function y= x^5 - 3x^3 -10x discover all real zeros. y = x(x^4 - 3x^2 - 10) = x(x^2 - 5)(x^2 + 2) real zeros are 0,+/- sqrt(5), (and +/- i sqrt(2) are complicated zeros) 4. The polynomial of degree 5, P(x) has maximum excellent coefficient a million, has roots of multiplicity 2 at x=2 and x=0 and a root of multiplicity a million at x= -a million hit upon a a hazard formula for P(x) multiplicities will become exponents on aspects P(x) = x^2(x - 2)^2* (x + a million) develop as needed... greater to stick to, yet you will possibly desire to be attempting those on your man or woman!

Since the Fundamental Law of Algebra states that there are four roots to this polynomial, the two other roots are the conjucate pair of 6 - 5i and -1 + √ 7, which is 6 + 5i and -1 - √ 7.

So the quartic polynomial is:

(x -6 +5i])(x -6 -5i)(x +1 -√7)(x +1 +√7)

=(x² -12x -61)(x² +2x +6)

=x^4 -10x³ -79x² -194x -366

This polynomial can be mutiplied be any rational factor, but the ratio of the coeffecients of the powers of x must be kept constant.

HOW did u get that sign before '7' up there?

6+5i, -1-sqrrt(7)

Please give me best answer thanks!