f(x)=x4+8x3+16x2
g(x)=5−4x2−16x
To find the point of intersection, let's equate both equations. This gives
x4+8x3+16x2=5−4x2−16x
x4+8x3+16x2−5+4x2+16x=0
x4+8x3+16x2+4x2+16x−5=0
x4+8x3+20x2+16x−5=0
x4+ 8x3+20x2+16x=5
x(x+2)2(x+4)=5
[x(x+4)−1][x(x+4)+5]=0
[x2+4x−1][x2+4x+5]=0
Hence, we set [x2+4x−1]=0and[x2+4x+5]=0
[x2+4x−1]=0
solving quadratically, we have
⟹x=2a−b±b2−4ac⟹x=2(1)−4±42−4(1)(−1)
⟹2−4±16+4
⟹2−4±20
⟹2−4±220
⟹−2±225
⟹−2±5
⟹−2+5or⟹−2−5
Also, solving the second Quadratic equation similarly, we have
[x2+4x+5]=0
⟹x=2(1)−4±42−4(1)(5)
⟹2−4±16−20
⟹2−4±−4
⟹2−4±2−4
⟹−2±22i
⟹−2±i
⟹−2+ior−2−i
Finding the roots of the polynomial,
x=(−5−2,−2+5,−2−i,−2+i)
let y=f(x)=g(x)
substituting x into either f(x)org(x) we get the same value for y.
Let's substitute x into g(x)
y=5−4x2−16x
⟹y=(1,25)
The points of intersection are
(−5−2,1),(−2+5,1),(−2−i,25),(−2+i,25).
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