Answer:
Yes, it is true that [tex]x^2+3x-1[/tex] is a factor of [tex]x^4+3x^3-2x^2-3x+1[/tex].
Step-by-step explanation:
Let us try to factorize [tex]x^4+3x^3-2x^2-3x+1[/tex]
[tex]x^4+3x^3-2x^2-3x+1\\\Rightarrow x^4-2x^2+1-3x+3x^3[/tex]
Let us try to make a whole square of the given terms:
[tex]\Rightarrow (x^2)^2-2\times x^2 \times 1+1^2+3x^3-3x\\\Rightarrow (x^2-1)^2+3x^3-3x\\[/tex]
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Formula used above:
[tex]a^{2} -2 \times a \times b +b^{2} = (a-b)^2[/tex]
In the above equation, we had [tex]a = x, b = 1[/tex].
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Further solving the above equation, taking [tex]3x[/tex] common out of [tex]3x^3-3x[/tex]
[tex]\Rightarrow (x^2-1)^2+3x(x^2-1)\\[/tex]
Taking [tex](x^{2} -1)[/tex] common out of the above term:
[tex]\Rightarrow (x^2-1)((x^2-1)+3x)\\\Rightarrow (x^2-1)(x^2+3x-1)[/tex]
So, the two factors are [tex](x^2-1)\ and\ (x^2+3x-1)[/tex].
[tex]\therefore[/tex] The statement that [tex]x^2+3x-1[/tex] is a factor of [tex]x^4+3x^3-2x^2-3x+1[/tex] is True.
