Solve the equation by factoring: 16x^(2)-30 x-2x^(3)=0 Answer: x=

Rewrite in Standard Form: First, let's rewrite the equation in standard form by arranging the terms in descending order of their exponents. The given equation is 16x^2 - 30x - 2x^3 = 0. Rearrange the terms to get -2x^3 + 16x^2 - 30x = 0.Factor Out Common Factor: Next, factor out the greatest common factor, which in this case is -2x. So, -2x(x^2 - 8x + 15) = 0.Factor Quadratic Expression: Now, we need to factor the quadratic expression x^2 - 8x + 15. To find factors of 15 that add up to -8, we can use -3 and -5. So, x^2 - 8x + 15 can be factored as (x - 3)(x - 5).Find Roots: The factored form of the original equation is now -2x(x - 3)(x - 5) = 0. To find the roots, we set each factor equal to zero and solve for x. -2x = 0, x - 3 = 0, and x - 5 = 0.Solving each equation for x gives us the roots: From -2x = 0, we get x = 0. From x - 3 = 0, we get x = 3. From x - 5 = 0, we get x = 5.

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Q. Solve the equation by factoring:

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16 x^{2}-30 x-2 x^{3}=0

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Answer:

x=

Rewrite in Standard Form: First, let's rewrite the equation in standard form by arranging the terms in descending order of their exponents. $\newline$ The given equation is $16x^2 - 30x - 2x^3 = 0$ . $\newline$ Rearrange the terms to get $-2x^3 + 16x^2 - 30x = 0$ .
Factor Out Common Factor: Next, factor out the greatest common factor, which in this case is $-2x$ . So, $-2x(x^2 - 8x + 15) = 0$ .
Factor Quadratic Expression: Now, we need to factor the quadratic expression $x^2 - 8x + 15$ . To find factors of $15$ that add up to $-8$ , we can use $-3$ and $-5$ . So, $x^2 - 8x + 15$ can be factored as $(x - 3)(x - 5)$ .
Find Roots: The factored form of the original equation is now $-2x(x - 3)(x - 5) = 0$ . To find the roots, we set each factor equal to zero and solve for $x$ . $-2x = 0$ , $x - 3 = 0$ , and $x - 5 = 0$ .
Find Roots: The factored form of the original equation is now $-2x(x - 3)(x - 5) = 0$ . $\newline$ To find the roots, we set each factor equal to zero and solve for $x$ . $\newline$ $-2x = 0$ , $x - 3 = 0$ , and $x - 5 = 0$ .Solving each equation for $x$ gives us the roots: $\newline$ From $-2x = 0$ , we get $x = 0$ . $\newline$ From $x - 3 = 0$ , we get $x = 3$ . $\newline$ From $x - 5 = 0$ , we get $x$ $1$ .