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The function below has at least one rational zero. Use this fact to find all zeros of the function. $h(x) = 5x^3 + 12x^2 + 12x + 7$ If there is more than one zero, separate them with co

          The function below has at least one rational zero.
Use this fact to find all zeros of the function.
$h(x) = 5x^3 + 12x^2 + 12x + 7$
If there is more than one zero, separate them with co
        
The function below has at least one rational zero.
Use this fact to find all zeros of the function.
h(x) = 5x^3 + 12x^2 + 12x + 7
If there is more than one zero, separate them with co

Added by Tracy C.

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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The function below has at least one rational zero. Use this fact to find all zeros of the function. h(x)=5x^(3)+12x^(2)+12x+7 If there is more than one zero, separate them with The function below has at least one rational zero Use this fact to finda/zeros of the function hx=5x+12x2+12x+7 If there is more than one zero, separate them with c
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The function below has at least one rational zero. Use this fact to find all zeros of the function. g(x) = 5x^4 - 6x^3 - 10x^2 + 3x + 2 If there is more than one zero, separate them with commas. Write exact values, not decimal approximations.

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Transcript

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00:01 So what i did with this problem is i went to my graphing calculators, typed in, so like right here, x equals n8 .1, my graph looked like this kind of go this way, and they tell you rational zero.
00:12 If i were guessing what this would be, based of the numbers they give, that x is equal to 6q plus x square, 12x5, is, first of all, that that is a negative number.
00:26 Know the rational root theorem, it has to be factors of the constant divided by factors of the coefficient.
00:31 So i'm going to guess negative five -sevenths.
00:36 And what that means is i would probably do a long division, just because a long division makes more sense to me.
00:48 So let's circle this and see if it actually works.
00:52 I like to multiply seven and then add five to the left side.
00:57 So if that really is 0, if i do long division, divide 6 cubed plus 12 squared, plus of x plus 5, i'll give you a remainder of 0.
01:08 Let's see.
01:10 6x squared gives me 7x cubed, but i'll multiply 5.
01:14 And then when i act down, the 1 i get 7x squared plus 12, just bringing down.
01:20 So times an x will give me 7x squared again, and then plus 5x...
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