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ArunSharma12
If a, b, and c are the roots of the equation \(x3–4x2+3x+5=0\), what is the value of (a+1)(b+1)(c+1)?

A. -3
B. -1
C. 0
D. 1
E. 3


for cubic equations, \(ax3 + bx2 +cx +d = 0\), with roots a,b and c

a+b+c = -b/a
ab+bc+ac = c/a
abc = -d/a

(a+1)(b+1)(c+1) = (a+b+c) + (ab+bc+ac) + abc + 1 = 4 + 3 -5 + 1 = 3

Ans: E

Can you please explain those three formulas?
how does a+b+c = -b/a?
1-4+3 = 4/1 -> that doesn't work?
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ArunSharma12
If a, b, and c are the roots of the equation \(x3–4x2+3x+5=0\), what is the value of (a+1)(b+1)(c+1)?

A. -3
B. -1
C. 0
D. 1
E. 3


for cubic equations, \(ax3 + bx2 +cx +d = 0\), with roots a,b and c

a+b+c = -b/a
ab+bc+ac = c/a
abc = -d/a

(a+1)(b+1)(c+1) = (a+b+c) + (ab+bc+ac) + abc + 1 = 4 + 3 -5 + 1 = 3

Ans: E

Can you please explain those three formulas?
how does a+b+c = -b/a?
1-4+3 = 4/1 -> that doesn't work?

you are right :thumbup: , I should be more careful with my choice of variables. I have edited my post.

these are the standard sum of the roots and product of roots formula for cubic equations, similar to what we got for quadratic equations.
sum of roots = -b/a product of roots = c/a

if you are looking for the proof -> https://brilliant.org/wiki/cubic-equations/
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a, b and c are the roots so the equation can be written as E(x) = (x-a)(x-b)(x-c).
So E(-1)= -1-4-3+5 = -3 = (-1-a)(-1-b)(-1-c) = - (a+1)(b+1)(c+1)
Then the solution is E. 3
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so I need to bother with this question?
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