Respuesta :
Answer:
1) Perfect square, 2) Perfect square, 3) Not a perfect square, 4) Perfect square, 5) Perfect square, 6) Not a perfect square, 7) Perfect square, 8) Not a perfect square, 9) Not a perfect square, 10) Not a perfect square.
Step-by-step explanation:
From Algebra we must remember that a trinomial is a perfect square if and only if:
[tex](a+b)^{2} = a^{2}+2\cdot a\cdot b + b^{2}[/tex], [tex]\forall \,a,b\in\mathbb{R}[/tex] (Eq. 1)
Now we proceed to prove each trinomial:
1) [tex]a^{2}+2\cdot a +1[/tex]:
If [tex]a^2 =a^2[/tex] and [tex]b^{2} = 1[/tex], then [tex]a = \pm a[/tex] and [tex]b =\pm 1[/tex]. Therefore, [tex]2\cdot a\cdot b = 2\cdot (\pm a)\cdot (\pm 1) = \pm 2\cdot a[/tex]. In a nutshell, we conclude that this polynomial is a perfect square.
2) [tex]4+4\cdot a +a^{2}[/tex]
If [tex]a^{2} = 4[/tex] and [tex]b^{2} = a^{2}[/tex], then [tex]a = \pm 2[/tex] and [tex]b = \pm a[/tex]. Therefore, [tex]2\cdot a \cdot b = 2\cdot (\pm 2)\cdot (\pm a) = \pm 4\cdot a[/tex]. In a nutshell, we conclude that this polynomial is a perfect square.
3) [tex]a^{2}-3\cdot a +9[/tex]
If [tex]a^{2} = a^{2}[/tex] and [tex]b^{2} = 9[/tex], then [tex]a = \pm a[/tex] and [tex]b = \pm 3[/tex]. Therefore, [tex]2\cdot a \cdot b = 2\cdot (\pm a)\cdot (\pm 3) = \pm 6\cdot a[/tex]. In a nutshell, we conclude that this polynomial is not a perfect square.
4) [tex]4\cdot a^{2}+24\cdot a + 36[/tex]
If [tex]a^{2} = 4\cdot a^{2}[/tex] and [tex]b^{2} = 36[/tex], then [tex]a = \pm 2\cdot a[/tex] and [tex]b = \pm 6[/tex]. Therefore, [tex]2\cdot a \cdot b = 2\cdot (\pm 2\cdot a)\cdot (\pm 6) = 24\cdot a[/tex]. In a nutshell, we conclude that this polynomial is a perfect square.
5) [tex]a^{4}-10\cdot a +25[/tex]
If [tex]a^{2} = a^{4}[/tex] and [tex]b^{2} = 25[/tex], then [tex]a = \pm a^{2}[/tex] and [tex]b = \pm 5[/tex]. Therefore, [tex]2\cdot a \cdot b = 2\cdot (\pm a^{2})\cdot (\pm 5) = \pm 10\cdot a^{2}[/tex]. In a nutshell, we conclude that this polynomial is a perfect square.
6) [tex]25\cdot a^{2}+20\cdot a + 9[/tex]
If [tex]a^{2} = 25\cdot a^{2}[/tex] and [tex]b = 9[/tex], then [tex]a = \pm 5\cdot a[/tex] and [tex]b = \pm 3[/tex]. Therefore, [tex]2\cdot a \cdot b = 2\cdot (\pm 5\cdot a)\cdot (\pm 3) = \pm 30\cdot a[/tex]. In a nutshell, we conclude that this polynomial is not a perfect square.
7) [tex]a^{2}\cdot b^{2}-14\cdot a \cdot b + 49[/tex]
If [tex]a^{2} = a^{2}\cdot b^{2}[/tex] and [tex]b^{2} = 49[/tex], then [tex]a = \pm a\cdot b[/tex] and [tex]b = \pm 7[/tex]. Therefore, [tex]2\cdot a\cdot b = 2\cdot (\pm a\cdot b)\cdot (\pm 7) = \pm 14\cdot a \cdot b[/tex]. In a nutshell, we conclude that this polynomial is a perfect square.
8) [tex]a^{2}-2\cdot a -1[/tex]
If [tex]a^{2} = a^{2}[/tex] and [tex]b^{2} = -1[/tex], then [tex]a = \pm a[/tex] and [tex]b = \pm i[/tex]. Therefore, [tex]2\cdot a \cdot b = \pm 2\cdot a \cdot i[/tex]. In a nutshell, we conclude that this polynomial is not a perfect square.
9) [tex]16\cdot a^{2} + 24\cdot a +36[/tex]
If [tex]a^{2} = 16\cdot a^{2}[/tex] and [tex]b^{2} = \pm 36[/tex], then [tex]a = \pm 4\cdot a[/tex] and [tex]b = \pm 6[/tex]. Therefore, [tex]2\cdot a \cdot b = 2\cdot (\pm 4\cdot a)\cdot (\pm 6) = \pm 48\cdot a[/tex]. In a nutshell, we conclude that this polynomial is not a perfect square.
10) [tex]10+6\cdot a +a^{2}[/tex]
If [tex]a^{2} = 10[/tex] and [tex]b^{2} = a^{2}[/tex], then [tex]a = \pm\sqrt{10}[/tex] and [tex]b = \pm a[/tex]. Therefore, [tex]2\cdot a \cdot b = \pm 2\cdot \sqrt{10}\cdot a[/tex]. In a nutshell, we conclude that this polynomial is not a perfect square.
