Question 177840
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The first thing to do is to simplify this horror by letting *[tex \Large t = x^2 +3x +1].


Now you can substitute giving you:


*[tex \Large t^2 - 4t - 5]


Since *[tex \Large -5 \times 1 = -5] and *[tex \Large -5 + 1 = -4], this quadratic polynomial in <i>t</i> factors quite tidily:


*[tex \Large (t - 5)(t + 1)]


Now replace <i>t</i> with the original expression:


*[tex \Large (x^2 +3x +1 - 5)(x^2 +3x +1 + 1)]


Collect terms:


*[tex \Large (x^2 +3x - 4)(x^2 +3x + 2)]


Now since *[tex \Large 4 \times (-1) = -4] and *[tex \Large 4 - 1 = 3] the first factor factors to *[tex \Large (x + 4)(x - 1)]


And since *[tex \Large 2 \times 1 = 2] and *[tex \Large 2 + 1 = 3] the second factor factors to *[tex \Large (x + 2)(x + 1)]


Putting it all together:


*[tex \Large (x^2 + 3x + 1)^2 - 4(x^2 + 3x + 1)-5 = (x + 4)(x - 1)(x + 2)(x + 1)]


And, as you might suspect, the graph shows four zeros that match the factors:

{{{drawing(
600,600,-5,5,-5,5,
grid(1),
graph(
600,600,-5,5,-5,5,
y = (x^2 + 3x + 1)^2 - 4(x^2 + 3x + 1)-5
))}}}


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