Can you show that $p(x)=2x^3 + 7x^2 + kx - k$ is the produce of 3 linear factors, 2 of which are identical? Show that $k$ can take 3 distinct values

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Answer 1 · 2017-04-04T11:47:00

See below.

Define $q(x)=a(x+b)^2(x+c)$ and consider

$p(x)=q(x)$ or

$p(x)=2x^3 + 7x^2 + kx - k = a(x+b)^2(x+c)$

Grouping coefficients we get

$(2-a)x^3+(7-(2ab+ac))x^2+(k-(a b^2 + 2 a b c))x-(k+ab^2c)=0$

Now solving for $a,b,c,k$

${(a=2),(2ab+ac=7),(a b^2 + 2 a b c=k),(ab^2c=-k):}$

we obtain

$((a,b,c,k),(2,-7/4,7,-343/8),(2,0,7/2,0),(2,2,-1/2,4))$

Can you show that p(x)=2x^3 + 7x^2 + kx - kp(x)=2x^3 + 7x^2 + kx - k is the produce of 3 linear factors, 2 of which are identical? Show that kk can take 3 distinct values