How can I factorize (y^(2)+3y+8)(y^(2)+3y-7)+54?
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You need to write the first factor `y^2 + 3y + 8` such that:
`y^2 + 3y + 8 = y^2 + 3y + 7 + 1`
`(y^2 + 3y + 7 + 1)(y^2 + 3y- 7) + 54`
`(y^2 + 3y + 7)(y^2 + 3y - 7) + y^2 + 3y - 7 + 54`
Converting the product `(y^2 + 3y + 7)(y^2 + 3y - 7)` into difference of two squares yields:
`(y^2 + 3y)^2 - 49+ y^2 + 3y - 7 + 54`
`(y^2 + 3y)^2 + (y^2 + 3y) - 2 = 0`
You need to use the following substitution such that:
`y^2 + 3y = t`
`t^2 + t - 2 = 0`
You should use quadratic formula such that:
`t_(1,2) = (-1+-sqrt(1+8))/2 => t_(1,2) = (-1+-3)/2`
`t_1 = 1 ; t_2 = -2`
You may write the factored form of the equation `t^2 + t - 2 = 0` such that:
`t^2 + t - 2 = (t - t_1)(t - t_2)`
`t^2 + t - 2 = (t - 1)(t + 2)`
You need to substitute back `y^2 + 3y` for t such that:
`(y^2 + 3y)^2 + (y^2 + 3y) - 2 = (y^2 + 3y - 1)(y^2 + 3y + 2)`
You should find the roots of equations `y^2 + 3y - 1 = 0` and `y^2 + 3y + 2 = 0` such that:
`y_(1,2) = (-3+-sqrt(9+4))/2 => y_(1,2) = (-3+-sqrt13)/2`
`y_(3,4) = (-3+-sqrt(9-8))/2 => y_(3,4) = (-3+-1)/2`
`y_3 = -1; y_2 = -2`
Hence, evaluating the factored form yields `(y^2 + 3y + 8)(y^2 + 3y - 7) + 54 = (y - (-3+sqrt13)/2)(y+ (3+sqrt13)/2)(y + 1)(y+2).`
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