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Answer :
To find the real zeros of the polynomial function [tex]\( f(x) = (x^2 - 36)(x^2 - 81) \)[/tex], we need to set each factor equal to zero and solve for [tex]\( x \)[/tex].
1. Start with the first factor: [tex]\( x^2 - 36 = 0 \)[/tex].
- Add 36 to both sides: [tex]\( x^2 = 36 \)[/tex].
- Take the square root of both sides: [tex]\( x = \pm 6 \)[/tex].
- This gives us two real zeros: [tex]\( x = 6 \)[/tex] and [tex]\( x = -6 \)[/tex].
2. Move to the second factor: [tex]\( x^2 - 81 = 0 \)[/tex].
- Add 81 to both sides: [tex]\( x^2 = 81 \)[/tex].
- Take the square root of both sides: [tex]\( x = \pm 9 \)[/tex].
- This provides two more real zeros: [tex]\( x = 9 \)[/tex] and [tex]\( x = -9 \)[/tex].
By solving both equations, we have found four real zeros for the polynomial function:
- [tex]\( x = -9 \)[/tex]
- [tex]\( x = -6 \)[/tex]
- [tex]\( x = 6 \)[/tex]
- [tex]\( x = 9 \)[/tex]
Therefore, the real zeros of the polynomial function are [tex]\( x = -9, x = -6, x = 6, \)[/tex] and [tex]\( x = 9 \)[/tex] only.
1. Start with the first factor: [tex]\( x^2 - 36 = 0 \)[/tex].
- Add 36 to both sides: [tex]\( x^2 = 36 \)[/tex].
- Take the square root of both sides: [tex]\( x = \pm 6 \)[/tex].
- This gives us two real zeros: [tex]\( x = 6 \)[/tex] and [tex]\( x = -6 \)[/tex].
2. Move to the second factor: [tex]\( x^2 - 81 = 0 \)[/tex].
- Add 81 to both sides: [tex]\( x^2 = 81 \)[/tex].
- Take the square root of both sides: [tex]\( x = \pm 9 \)[/tex].
- This provides two more real zeros: [tex]\( x = 9 \)[/tex] and [tex]\( x = -9 \)[/tex].
By solving both equations, we have found four real zeros for the polynomial function:
- [tex]\( x = -9 \)[/tex]
- [tex]\( x = -6 \)[/tex]
- [tex]\( x = 6 \)[/tex]
- [tex]\( x = 9 \)[/tex]
Therefore, the real zeros of the polynomial function are [tex]\( x = -9, x = -6, x = 6, \)[/tex] and [tex]\( x = 9 \)[/tex] only.
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