Question

factor the polynomial expression $x^4 + 18x^2 + 81$ to determine the missing terms. drag the expressions to the correct locations on the image. not all expressions will be used. $x^4 + 18x^2 + 81 = (\quad)^2 + 2(\quad)x^2 + ((\quad)^2)^2$ $x^4 + 18x^2 + 81 = (\quad)^2$ $x^2 + 9$ $x^2 - 9$ $x^2$ $3^2$ $3$ $81$ $x$

Explanation:

Step1: Identify the perfect square form

Recall the perfect square trinomial formula: \(a^2 + 2ab + b^2=(a + b)^2\). For the polynomial \(x^4+18x^2 + 81\), we can rewrite \(x^4=(x^2)^2\) and \(81 = 9^2=(3^2)^2\). Also, \(18x^2=2\times x^2\times9\). So, \(x^4+18x^2 + 81=(x^2)^2+2(x^2)(9)+(9)^2\), but since \(9 = 3^2\), we can also write the middle term as \(2(x^2)(3^2)\) and the last term as \(((3)^2)^2\) in the first given structure. So the first blank (for \((\quad)^2\)) is \(x^2\), the second blank (for \(2(\quad)x^2\)) is \(9\) (or \(3^2\), but \(9\) is simpler, and also we can see that \(9 = 3^2\), so the third blank (for \((( \quad)^2)^2\)) is \(3\)).

Step2: Factor the perfect square trinomial

Using the perfect square trinomial formula \(a^2+2ab + b^2=(a + b)^2\), where \(a=x^2\) and \(b = 9\) (since \(a=x^2\), \(b = 9\) as \(2ab=2\times x^2\times9 = 18x^2\)), so \(x^4+18x^2+81=(x^2 + 9)^2\).

Answer:

First structure: \(x^4 + 18x^2 + 81 = (\boldsymbol{x^2})^2 + 2(\boldsymbol{9})x^2 + ((\boldsymbol{3})^2)^2\) (or for the second blank, it can also be \(3^2\) but \(9\) is more direct, and third blank is \(3\))
Second structure: \(x^4 + 18x^2 + 81 = (\boldsymbol{x^2 + 9})^2\)

(For the drag - and - drop, the first blank: \(x^2\), second blank: \(9\) (or \(3^2\)), third blank: \(3\), and the last blank: \(x^2 + 9\))