How do you factor \(80y^2 - 4y - 4\), or is it prime?

0\\\\y_1=\frac{-b-\sqrt\Delta}{2a};\ y_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{1296}=36\\\\y_1=\frac{4-36}{2\cdot80}=\frac{-32}{160}=-\frac{1}{5};\ y_2=\frac{4+36}{2\cdot80}=\frac{40}{160}=\frac{1}{4}"> 80 y 2 − 4 y − 4 a = 80 ; b = − 4 ; c = − 4 Δ = b 2 − 4 a c Δ = ( − 4 ) 2 − 4 ⋅ 80 ⋅ ( − 4 ) = 16 + 1280 = 1296 > 0 y 1 ​ = 2 a − b − Δ ​ ​ ; y 2 ​ = 2 a − b + Δ ​ ​ Δ ​ = 1296 ​ = 36 y 1 ​ = 2 ⋅ 80 4 − 36 ​ = 160 − 32 ​ = − 5 1 ​ ; y 2 ​ = 2 ⋅ 80 4 + 36 ​ = 160 40 ​ = 4 1 ​
80 y 2 − 4 y − 4 = 80 ( y + 5 1 ​ ) ( y − 4 1 ​ ) 80 ( y + 5 1 ​ ) ( y − 4 1 ​ ) = 4 ⋅ 5 ⋅ 4 ⋅ ( y + 5 1 ​ ) ( y − 4 1 ​ ) = 4 ( 5 y + 1 ) ( 4 y − 1 )
80 y 2 − 4 y − 4 = 80 y 2 − 4 y − 16 y + 16 y − 4 = 80 y 2 − 20 y + 16 y − 4 = 20 y ( 4 y − 1 ) + 4 ( 4 y − 1 ) = ( 4 y − 1 ) ( 20 y + 4 )

Answered by Anonymous | 2024-06-10

80 y 2 − 4 y − 4 = 4 ( 20 y 2 − 5 y + 4 y − 1 ) = 4 [ 5 y ( 4 y − 1 ) − ( 4 y − 1 )] = = 4 ( 4 y − 1 ) ( 5 y − 1 )

Answered by kate200468 | 2024-06-10

To factor the expression 80 y 2 − 4 y − 4 , we calculated the discriminant, found the roots using the quadratic formula, and expressed it in the factored form as 4 ( 5 y + 1 ) ( 4 y − 1 ) . The expression is not prime, as it can be factored into these simpler components. This shows that it has real roots and can be represented as a product of linear factors.
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Answered by Anonymous | 2024-12-20