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A polynomial P is given. P(x) = x^4 - 5x^3 + 6x^2 + 4x - 8 (a) Find all the real zeros of P. (Enter your answers as a comr x = 

          A polynomial P is given.
P(x) = x^4 - 5x^3 + 6x^2 + 4x - 8
(a) Find all the real zeros of P. (Enter your answers as a comr
x =
A polynomial P is given. P(x) = x^4 - 5x^3 + 6x^2 + 4x - 8 (a) Find all the real zeros of P. (Enter your answers as a comr x =

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Elementary and Intermediate Algebra
Elementary and Intermediate Algebra
Alan S. Tussy, R. David Gustafson 5th Edition
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A polynomial P is given. P(x)=x^(4)-5x^(3)+6x^(2)+4x-8 (a) Find all the real zeros of P. (Enter your answers as a comr x= A polynomial P is given P()=x4 -5x3+6x2+4x -8 (a) Find all the real zeros of P.(Enter your answers as a com X=
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Transcript

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00:01 Okay, here we have p of x is equal to x -6 minus 7 x cubed minus 8.
00:06 Now to find a 0, take p of x is equal to 0.
00:11 So your middle term is negative 7x cube.
00:17 So we can write your negative 8x cube plus x -cube negative negative 8 is equal to 0.
00:25 Now, factor out x -cube for the first two terms.
00:29 So your x -cube multiplied by x -cube.
00:32 Minus 8 and vector out 1 for the last 2 terms so here 1 multiplied by x cube minus 8 now here we can factor out x cube minus 8 and here x cube minus 8 multiplied by x q plus 1 is equal to 0 right so here we have 8 so you can write x cube minus 2 q and for the for the second term we can write x cubed cube plus one cube because your one cube is equal to one right okay for for this term we have to use this first formula that is x minus 2 multiplied by x square plus a b that is 2 multiply by x that is 2x and plus b square so here 2 to 2 that is equal to 4 and for x cube plus 1 cube we have to use this second formula so here x plus 1 multiply by x squared minus 1 into x that is equal to x and here one square is equal to 1 right okay here x minus 2 is equal to 0 so here we can write x is equal to 2 and here x plus 1 is equal to 0 so we can write here x is equal to minus 1.
01:58 Now the remaining terms are this and this right so here x square plus 2x plus 4 is equal to 0 and x squared minus x plus 1 is equal to 0 now we have to use the codetic formula to solve both this equation so our codetic formula that is equal to x is equal to negative b plus or minus b square minus 4a c divided by 2a okay so here b is equal to 2 so here x is equal to negative 2 plus or minus here b squared so here 2 rest 2 that is equal to 4 here a is equal to 1 and c is equal to 4 here we have 2 a so here a is equal to 1 so 2 multiply by 1 that is equal to 2 here 4 minus 4 multiply by 4 that is equal to 16 here 4 minus 16 that is equal to negative 12 and here's square root of negative 12 that is equal to 2 multiply by square root of 3i.
03:29 Here 2 divided by 2, that is 1 and here 2 and 2 will be cancelled out...
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