Mathematics, 09.11.2020 17:00 granthazenp5e9mj

Determine whether each trinomial is a perfect square or not.. 1.A2+2a+1 2.4+4a+a2 3.A2-3a+9 4.4a2+24a+36 5.A4-10a+25 6.25a2+20a+9 7.A2b2-14ab+49 8.A2-2a-1 9.16a2+24a+36 10.10+6a+a2

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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:

$(a+b)^{2} = a^{2}+2\cdot a\cdot b + b^{2}$ , $\forall \,a,b\in\mathbb{R}$ (Eq. 1)

Now we proceed to prove each trinomial:

1) $a^{2}+2\cdot a +1$ :

If a^2 =a^2 and $b^{2} = 1$ , then $a = \pm a$ and $b =\pm 1$ . Therefore, $2\cdot a\cdot b = 2\cdot (\pm a)\cdot (\pm 1) = \pm 2\cdot a$ . In a nutshell, we conclude that this polynomial is a perfect square.

2) $4+4\cdot a +a^{2}$

If $a^{2} = 4$ and $b^{2} = a^{2}$ , then $a = \pm 2$ and $b = \pm a$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm 2)\cdot (\pm a) = \pm 4\cdot a$ . In a nutshell, we conclude that this polynomial is a perfect square.

3) $a^{2}-3\cdot a +9$

If $a^{2} = a^{2}$ and $b^{2} = 9$ , then $a = \pm a$ and $b = \pm 3$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm a)\cdot (\pm 3) = \pm 6\cdot a$ . In a nutshell, we conclude that this polynomial is not a perfect square.

4) $4\cdot a^{2}+24\cdot a + 36$

If $a^{2} = 4\cdot a^{2}$ and $b^{2} = 36$ , then $a = \pm 2\cdot a$ and $b = \pm 6$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm 2\cdot a)\cdot (\pm 6) = 24\cdot a$ . In a nutshell, we conclude that this polynomial is a perfect square.

5) $a^{4}-10\cdot a +25$

If $a^{2} = a^{4}$ and $b^{2} = 25$ , then $a = \pm a^{2}$ and $b = \pm 5$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm a^{2})\cdot (\pm 5) = \pm 10\cdot a^{2}$ . In a nutshell, we conclude that this polynomial is a perfect square.

6) $25\cdot a^{2}+20\cdot a + 9$

If $a^{2} = 25\cdot a^{2}$ and b = 9 , then $a = \pm 5\cdot a$ and $b = \pm 3$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm 5\cdot a)\cdot (\pm 3) = \pm 30\cdot a$ . In a nutshell, we conclude that this polynomial is not a perfect square.

7) $a^{2}\cdot b^{2}-14\cdot a \cdot b + 49$

If $a^{2} = a^{2}\cdot b^{2}$ and $b^{2} = 49$ , then $a = \pm a\cdot b$ and $b = \pm 7$ . Therefore, $2\cdot a\cdot b = 2\cdot (\pm a\cdot b)\cdot (\pm 7) = \pm 14\cdot a \cdot b$ . In a nutshell, we conclude that this polynomial is a perfect square.

8) $a^{2}-2\cdot a -1$

If $a^{2} = a^{2}$ and $b^{2} = -1$ , then $a = \pm a$ and $b = \pm i$ . Therefore, $2\cdot a \cdot b = \pm 2\cdot a \cdot i$ . In a nutshell, we conclude that this polynomial is not a perfect square.

9) $16\cdot a^{2} + 24\cdot a +36$

If $a^{2} = 16\cdot a^{2}$ and $b^{2} = \pm 36$ , then $a = \pm 4\cdot a$ and $b = \pm 6$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm 4\cdot a)\cdot (\pm 6) = \pm 48\cdot a$ . In a nutshell, we conclude that this polynomial is not a perfect square.

10) $10+6\cdot a +a^{2}$

If $a^{2} = 10$ and $b^{2} = a^{2}$ , then $a = \pm\sqrt{10}$ and $b = \pm a$ . Therefore, $2\cdot a \cdot b = \pm 2\cdot \sqrt{10}\cdot a$ . In a nutshell, we conclude that this polynomial is not a perfect square.

Answer from: Quest

Photo math will give you the step by step answer : )

Answer from: Quest

aloha~! my name is zalgo and i am here to you out on this amazing day. to answer your question, yes, nonzero numbers can have a greatest common factor and that would be 1. still don't get it? let me considering every number is divisible by 1 and hence every number has 1 as a factor of theirs. so, if 2 nonzero numbers didn't share any gcf (greatest common factor), they would still have a factor in common and that would be 1.

i hope that this ! : p

"stay and stay proud! " - zalgo

(by the way, do you mind marking me as brainliest? i'd greatly appreciate it! mahalo! x3)

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Determine whether each trinomial is a perfect square or not.. 1.A2+2a+1 2.4+4a+a2 3.A2-3a+9 4.4a2+24...

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