# Calculate. y=3x^5 +5x^4 -2x^3

Calculate. $y=3{x}^{5}+5{x}^{4}-2{x}^{3}$
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Brenton Gay
X intercepts coorespond to the x coordinates where y = 0. Setting the function equal to 0
$3{x}^{5}+5{x}^{4}-2{x}^{3}=0$
Factoring out x cubed
${x}^{3}\left(3{x}^{2}+5x-2\right)=0$
${x}^{3}\left(3x-1\right)\left(x+2\right)=0$
the whole thing will be zero if any of the above factors are zero
${x}^{3}=0$
x=0
3x-1=0
3x=1
$x=\frac{1}{3}$
x+2=0
x=-2
So the x intercepts are -2, 0, 1/3

Bruno Thompson
Substitute $z={x}^{4}$. Then, the equation reads $y=3{z}^{2}+5z-2$.
Using the Pythagorean theorem ($z=\left[-b±\sqrt{{b}^{2}-4ac}\right]/2a$, where a = 3, b = 5, and c = -2: $z=\left[-5±\sqrt{52-\left(4\ast 3\ast -2\right)}\right]/\left(2\ast 3\right)=\left[-5±\sqrt{25+24}\right]/6=\left[-5±\sqrt{49}\right]/6=\left[-5±7\right]/6$ produces two solutions: 2/6 = 1/3 OR -12/6 = -2.
Therefore, the two z-intercepts of the substituted equation are z =1/3 and z = -2.
However, since we substituted $z={x}^{4}$ earlier, now to find our x-intercepts, we have . Thus, our solutions are x= 1/81 and x = 16.