Determine the zeros of f(x) = x4 – x3 + 7x2 – 9x – 18. help :))
x=-1,2
cheat
find them instantly, since you are already on line, just type it in http://www.wolframalpha.com/input/?i=x4+%E2%80%93+x3+%2B+7x2+%E2%80%93+9x+%E2%80%93+18.
how did you get tht DASHNI?
We'll have to substitute and check . Let's put x=-1 f(-1)=1+1+7+9-18=0 so x+1 is a factor, we'll create this factor and will find other factors f(x)=x^4-x^3+7x^2-9x-18 writing x^4 as x^3(x+1) this has added an extra x^3, so will subtract this by -x^3 x^4---->x^3(x+1)-x^3 so creating this factor in each term f(x)=x^3(x+1)-2x^2(x+1)+9x(x+1)-18x-18 f(x)=x^3(x+1)-2x^2(x+1)+9x(x+1)-18(x+1) f(x)=(x+1)(x^3-2x^2+9x-18)
now we'll find factors of (x^3-2x^2+9x-18) let's check if x=-1 is a factor of this -1-2-9-18 =-30 not equal to zero, let's check if x=2 is a factor 8-8+18-18=0 so x-2 is a factor now we'll create this factor in each term, like done before x^2(x-2)+9(x-2) (x-2)(x^2+9) so f(x)= (x+1)(x-2) (x^2+9) so the two real roots are x=-1, 2 the complex roots are \[x= \pm 3i\]
thank you very much:)
welcome:)
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