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Given that (x - 2) is a factor of x³ + x² - 5ax + 2a², 'a' can take the value of 2 or which other number?​

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well now, let's recall the remainder theorem, check your textbook since we know you're Da Bomb and love to check that theorem.

well, we know a factor is (x-2) for the polynomial f(x), that means that f(2) = 0, so let's use that


f(x)=x^3+x^2-5ax+2a^2\hspace{5em}\stackrel{\textit{a factor of f(x)}}{(x-2)}\qquad thus\qquad f(2)=0 \\\\\\ 0=(2)^3+(2)^2-5a(2)+2a^2\implies 0=8+4-10a+2a^2 \\\\\\ 0=2a^2-10a+12\implies 0=2(a^2-5a+6) \\\\\\ 0=a^2-5a+6\implies 0=(a-2)(a-3)\implies a= \begin{cases} 2\\ 3 \end{cases}

User Thewooster
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