STEP-BY-STEP SOLUTION:
= a^3 - 4a^2 + 12 - 3a
= a^2 ( a - 4 ) + 3 ( 4 - a )
= a^2 ( a - 4 ) - 3 ( a - 4 )
= ( a - 4 ) ( a^2 - 3 )
Please mark as brainliest if you found this helpful! : )
Thank you <3
This is my interpretation of bias, something that screws your acceptance of other people's view, and taking <span>into account only one view point. </span>
first off let's recall that a square has all equal sides, so its area is just one side squared, namely A = s², or A = (s)(s).
we know the area is 16j² + 24j + 9, that simply means that two twin factors are in it, and it also means that the area polynomial is a perfect square trinomial.
![\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2 \\\\[-0.35em] ~\dotfill\\\\ 16j^2+24j+9\implies 4^2j^2+2(4j)(3)+3^2\implies (4j)^2+2(4j)(3)+3^2 \\\\\\ (4j+3)^2\implies \stackrel{\textit{area}}{(4j+3)(4j+3)}~\hspace{7em} \stackrel{\textit{one side}}{4j+3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7Bperfect%20square%20trinomial%7D%20%5C%5C%5C%5C%20%28a%5Cpm%20b%29%5E2%5Cimplies%20a%5E2%5Cpm%20%5Cstackrel%7B%5Cstackrel%7B%5Ctext%7B%5Csmall%202%7D%5Ccdot%20%5Csqrt%7B%5Ctextit%7B%5Csmall%20a%7D%5E2%7D%5Ccdot%20%5Csqrt%7B%5Ctextit%7B%5Csmall%20b%7D%5E2%7D%7D%7B%5Cdownarrow%20%7D%7D%7B2ab%7D%20%2B%20b%5E2%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%2016j%5E2%2B24j%2B9%5Cimplies%204%5E2j%5E2%2B2%284j%29%283%29%2B3%5E2%5Cimplies%20%284j%29%5E2%2B2%284j%29%283%29%2B3%5E2%20%5C%5C%5C%5C%5C%5C%20%284j%2B3%29%5E2%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Barea%7D%7D%7B%284j%2B3%29%284j%2B3%29%7D~%5Chspace%7B7em%7D%20%5Cstackrel%7B%5Ctextit%7Bone%20side%7D%7D%7B4j%2B3%7D)
I hope I can show the image, but I hope this may help you, the answer is:
The center of the inscribed circle of ΔABC is <u>point S</u> , and the center of the circumscribed circle of ΔABC is <u>point P.</u>
