Question

what is the zeros of polynomial p (x)=(a²+b²)x+(a-b)² + (a+b)² ?​

Answer 1

$\huge\sf\mathbb\color{lightgreen} \underline{\colorbox{pink}{☯SoLuTiOn☯}}$

Step-by-step explanation:

$p(x) \: : \: ( {a}^{2} + {b}^{2} )x + {(a + b)}^{2} + {(a + b)}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: = {a}^{2} x + {b}^{2} x + {a}^{2} + {b}^{2} - 2ab + {a}^{2} + {b}^{2} + 2ab$

$\: \: \: \: \: \: \: \: \: \: \: \: = {a}^{2} x + {b}^{2} x + 2 {a}^{2} + {2b}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: = (x + 2)( {a}^{2} + {b}^{2} )$

$\: \: \: \: \: \: \: \: \: \: \: \: = x + 2 = 0$

$\huge\sf\mathbb\color{lightblue} \underline{\colorbox{r}{ \: \: \: \: \: \: \: \: \: \: \: \: \: x =( - 2) }}$

Answer 2

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$\longmapsto \sf p(x) \: : \: ( {a}^{2} + {b}^{2} )x + {(a + b)}^{2} + {(a + b)}^{2}$

$\longmapsto \sf{a}^{2} x + {b}^{2} x + {a}^{2} + {b}^{2} - 2ab + {a}^{2} + {b}^{2} + 2ab$

$\longmapsto \sf {a}^{2} x + {b}^{2} x + 2 {a}^{2} + {2b}^{2}$

$\longmapsto \sf(x + 2)( {a}^{2} + {b}^{2} )$

$\longmapsto \sf x + 2 = 0$

$\longmapsto\sf\mathbb\color{lightblue} \underline{\colorbox{r}{ x =( - 2) }}$