How do you find the zeroes of $f (x) = 10 x^{5} + 2 x^{4} - 75 x^{3} - 15 x^{2} + 125 x + 25$ $f (x) =10x^5 + 2x^4 − 75x^3 − 15x^2 + 125x + 25$ ?

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How do you find the zeroes of $f (x) = 10 x^{5} + 2 x^{4} - 75 x^{3} - 15 x^{2} + 125 x + 25$ $f (x) =10x^5 + 2x^4 − 75x^3 − 15x^2 + 125x + 25$ ?

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Answer 1 · 2015-11-27T22:34:22

Factor by grouping twice, then use the difference of squares identity twice and hence find zeros:

$\pm \frac{\sqrt{10}}{2}$ $+-sqrt(10)/2$ , $\pm \sqrt{5}$ $+-sqrt(5)$ , $- \frac{1}{5}$ $-1/5$

Explanation:

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

Factor by grouping:

$f (x) = 10 x^{5} + 2 x^{4} - 75 x^{3} - 15 x^{2} + 125 x + 25$ $f(x) = 10x^5+2x^4-75x^3-15x^2+125x+25$

$= (10 x^{5} + 2 x^{4}) - (75 x^{3} + 15 x^{2}) + (125 x + 25)$ $= (10x^5+2x^4)-(75x^3+15x^2)+(125x+25)$

$= 2 x^{4} (5 x + 1) - 15 x^{2} (5 x + 1) + 25 (5 x + 1)$ $= 2x^4(5x+1)-15x^2(5x+1)+25(5x+1)$

$= (2 x^{4} - 15 x^{2} + 25) (5 x + 1)$ $= (2x^4-15x^2+25)(5x+1)$

$= (2 x^{4} - 10 x^{2} - 5 x^{2} + 25) (5 x + 1)$ $= (2x^4-10x^2-5x^2+25)(5x+1)$

$= ((2 x^{4} - 10 x^{2}) - (5 x^{2} - 25)) (5 x + 1)$ $= ((2x^4-10x^2)-(5x^2-25))(5x+1)$

$= (2 x^{2} (x^{2} - 5) - 5 (x^{2} - 5)) (5 x + 1)$ $= (2x^2(x^2-5)-5(x^2-5))(5x+1)$

$= (2 x^{2} - 5) (x^{2} - 5) (5 x + 1)$ $= (2x^2-5)(x^2-5)(5x+1)$

$= ({(\sqrt{2} x)}^{2} - {(\sqrt{5})}^{2}) (x^{2} - {(\sqrt{5})}^{2}) (5 x + 1)$ $= ((sqrt(2)x)^2-(sqrt(5))^2)(x^2-(sqrt(5))^2)(5x+1)$

$= (\sqrt{2} x - \sqrt{5}) (\sqrt{2} x + \sqrt{5}) (x - \sqrt{5}) (x + \sqrt{5}) (5 x + 1)$ $= (sqrt(2)x-sqrt(5))(sqrt(2)x+sqrt(5))(x-sqrt(5))(x+sqrt(5))(5x+1)$

Hence $f (x)$ $f(x)$ has zeros for the following values of $x$ $x$

$\pm \frac{\sqrt{5}}{\sqrt{2}} = \pm \frac{\sqrt{10}}{2}$ $+-sqrt(5)/sqrt(2) = +-sqrt(10)/2$

$\pm \sqrt{5}$ $+-sqrt(5)$

$- \frac{1}{5}$ $-1/5$

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How do you find the zeroes of $f (x) = 10 x^{5} + 2 x^{4} - 75 x^{3} - 15 x^{2} + 125 x + 25$ $f (x) =10x^5 + 2x^4 − 75x^3 − 15x^2 + 125x + 25$ ?

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How do you find the zeroes of f (x) =10x^5 + 2x^4 − 75x^3 − 15x^2 + 125x + 25f(x)=10x5+2x4−75x3−15x2+125x+25 f (x) =10x^5 + 2x^4 − 75x^3 − 15x^2 + 125x + 25?

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Explanation:

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How do you find the zeroes of $f (x) = 10 x^{5} + 2 x^{4} - 75 x^{3} - 15 x^{2} + 125 x + 25$ $f (x) =10x^5 + 2x^4 − 75x^3 − 15x^2 + 125x + 25$ ?