Find k such that g(x)=3x^3+kx^2-kx+4 has x-2 as a factor.Solve in the real number system x^4-x^3+2x^2-4x-=0?

list potential rational zeros for second problem. Find a polynomial of degree 4 whose roots are -1,2i,3-i. Find all complex and real zeros(roots)of f(x)=x^3+13x^2+57x+85.


The functionf(x)=3x^4-4x^2-3 has a zero( root)between x=1 and x=1.5.Use bisection methodto find a root accurate to 2 decimal places.


List possible rational zeros of f(x)=4x^5+3x^4-13x^3+28


Given 2i is a zero of the function f(x)=x^4-6x^3+12x^2-24+32,Find the remaining zeros


Please help me solve these Thanks

Find k such that g(x)=3x^3+kx^2-kx+4 has x-2 as a factor.Solve in the real number system x^4-x^3+2x^2-4x-=0?
Find k such that g(x)=3x^3+kx^2-kx+4 has x-2 as a factor.





Use synthetic division:


2 | 3 k -k 4


| 0 6 2k+12 2k+24


3 k+6 k+12 2k+28





Thus, 2k + 28 = 0, which means that k = -14.





***





Solve in the real number system x^4-x^3+2x^2-4x-=0? list potential rational zeros





Potential rational zeros are those which can be expressed as a factor of last term divided by a factor of the leading term. Unfortunately, you left off the last term, and thus this problem makes little sense as presented.





***





Find a polynomial of degree 4 whose roots are -1,2i,3-i.





(x + 1)(x - 2i)(x - 3 + i)(x - 3 - i)


= (x + 1)(x² - 3x + ix - 2ix + 6i + 2)(x - 3 - i)


= (x + 1)(x² - 3x - ix + 6i + 2)(x - 3 - i)


= (x³ - 3x² - ix² + 6ix + 2x + x² - 3x - ix + 6i + 2)(x - 3 - i)


= (x³ - 2x² - ix² + 5ix - x + 6i + 2)(x - 3 - i)


= (x³ - 2x² - ix² + 5ix - x + 6i + 2)x - 3(x³ - 2x² - ix² + 5ix - x + 6i + 2) - i(x³ - 2x² - ix² + 5ix - x + 6i + 2)


= x^4 - 2x³ - ix³ + 5ix² - x² + 6ix + 2x - 3x³ + 6x² + 3ix² - 15ix + 3x - 18i - 6 - ix³ + 2ix² - x² + 5x + ix + 6 - 2i


= x^4 - 5x³ - 2ix³ + 10ix² + 4x² + 10x - 8ix - 20i


= x^4 - (5+2i)x³ + (4+10i)x² + (10-8i)x - 20i





***





Find all complex and real zeros(roots)of f(x)=x^3+13x^2+57x+85.





Judging from the graph, the only real root is x=-5.





Using synthetic division:


-5 | 1 13 57 85


| 0 -5 -40 -85


1 8 17 0





Thus,


x^3+13x^2+57x+85


= (x+5)(x^2 + 8x + 17)





Applying the quadratic equation to second factor:


x = {-8 ± √[(8)² - 4(1)(17)]}/[2(1)]


= [-8 ± √(64 - 68)]/2


= (-8 ± √-4)/2


= (-8 ± 2√-1)/2


= -4 ± √-1


= -4±i





***





The functionf(x)=3x^4-4x^2-3 has a zero( root)between x=1 and x=1.5.Use bisection methodto find a root accurate to 2 decimal places.





f(1) = -4


f(1.5) = 3.1875


f(1.25) = -1.925781


f(1.375) ≈ .1608887


f(1.3125) ≈ -.9879913


f(1.34375) ≈ -.4413881


f(1.359375) ≈ -.147394


f(1.3671875) ≈ .004937898


f(1.36328125) ≈ -.07167748


f(1.365234375) ≈ -.03348251





Thus, the zero in question is near x≈1.37





***





List possible rational zeros of f(x)=4x^5+3x^4-13x^3+28





As mentioned previously, potential rational zeros are those which can be expressed as a factor of last term divided by a factor of the leading term.





28 = 28·1 = 14·2 = 7·4 ⇒ 1, 2, 4, 7, 14, 28


4 = 4·1 = 2·2 ⇒ 1, 2, 4





1, ½, ¼, 2, 4, 7, 7/2, 7/4, 14, 28, -1, -½, -¼, -2, -4, -7, -7/2, -7/4, -14, -28





***





Given 2i is a zero of the function f(x)=x^4-6x^3+12x^2-24+32, Find the remaining zeros





Right off the bat, -2i is one, because the coefficients are all integers.





Using synthetic division:





2i | 1 -6 12 -24 32


| 0 2i -4-12i 24+16i -32


1 -6+2i 8-12i 16i 0





Again,





-2i | 1 -6+2i 8-12i 16i


| 0 -2i 12i -16i


1 -6 8 0





Thus, x^4-6x^3+12x^2-24+32 = (x - 2i)(x + 2i)(x² - 6x + 8)





Using the quadratic equation:


x² - 6x + 8 = 0


x = {-(-6) ± √[(-6)² - 4(1)(8)]}/[2(1)]


= [6 ± √(36 - 32)]/2


= (6 ± √4)/2


= (6 ± 2)/2


= 3 ± 1


= 4 and 2