Find all zeros of the function and write the polynomial as a product of linear factors.

\[ f(x) = x^4 + 29x^2 + 100 \]

f ( x ) = x 4 + 29 x 2 + 100 = ( x 2 ) 2 + 29 x 2 + 100 x 2 = v ; v ≥ 0 f ( v ) = v 2 + 29 v + 100 a = 1 ; b = 29 ; c = 100 Δ = b 2 − 4 a c ; v 1 ​ = 2 a − b − Δ ​ ​ ; v 2 ​ = 2 a − b + Δ ​ ​ Δ = 2 9 2 − 4 ⋅ 1 ⋅ 100 = 841 − 400 = 441 ; Δ ​ = 441 ​ = 21 v 1 ​ = 2 ⋅ 1 − 29 − 21 ​ = 2 − 40 ​ = − 20 < 0 ; v 2 ​ = 2 ⋅ 1 − 29 + 21 ​ = 2 − 8 ​ = − 4 < 0
F u n c t i o n ha s n o zeros . f ( x ) = x 4 + 29 x 2 + 100 = x 4 + 25 x 2 + 4 x 2 + 100 = x 2 ( x 2 + 25 ) + 4 ( x 2 + 25 ) = ( x 2 + 25 ) ( x 2 + 4 ) i s n o t im p oss ib l e w r i t e t h e p o l y n o mia \l a s a p ro d u c t o f l in e a r f a c t ors .

Answered by Anonymous | 2024-06-10

take x² = t t²+29t+100 =(t²+25t) +(4t+100) =t(t+25) + 4(t+25) =(t+4)(t+25) =(x²+4)(x²+25)

Answered by Shashank | 2024-06-10

The function has no real zeros, and can be factored into the product of linear factors that represent its complex roots. The polynomial can be expressed as f ( x ) = ( x − 5 i ) ( x + 5 i ) ( x − 2 i ) ( x + 2 i ) . Thus, all zeros of the function are complex: 5i, -5i, 2i, and -2i.
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Answered by Anonymous | 2024-10-10