Negate the Roots
The roots of the monic polynomial x 5 + a x 4 + b x 3 + c x 2 + d x + e are − r 1 , − r 2 , − r 3 , − r 4 , and − r 5 , where r 1 , r 2 , r 3 , r 4 , and r 5 are the roots of the polynomial x 5 + 9 x 4 + 1 3 x 3 − 5 7 x 2 − 8 6 x + 1 2 0 . Find ∣ a + b + c + d + e ∣ .
Details and assumptions
A root of a polynomial is a number where the polynomial is zero. For example, 6 is a root of the polynomial 2 x − 1 2 .
A polynomial is monic if its leading coefficient is 1. For example, the polynomial x 3 + 3 x − 5 is monic but the polynomial − x 4 + 2 x 3 − 6 is not.
The notation ∣ ⋅ ∣ denotes the absolute value. The function is given by ∣ x ∣ = { x − x x ≥ 0 x < 0 For example, ∣ 3 ∣ = 3 , ∣ − 2 ∣ = 2 .
The answer is 145.
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3 solutions
When (x - r)(x - s) = 0 is multiplied we got x^2 - (r+s)x + (rs) = 0, where r, s are roots. When (x - r)(x - s)(x - t) = 0 is multiplied we got x^3 -(r+s+t)x^2 + (rs + st + rt)x - (rst) = 0, where r,s and t are roots.
Noticing the pattern, we can say that: (x - r)(x - s)(x - t)(x - u)(x - v) = 0 is x^5 - (r+s+t+u+v)x^4 + (rs+rt+ru+rv+st+su+sv+tu+tv+uv)x^3 - (rst + rsu + rsv + rtu + rtv + ruv + stu + stv + suv + tuv)x^2 + (rstu + rstv + rtuv + stuv + rsuv)x - (rstuv) = 0. (or we can simply used combination at this point)
changing r to -r , s to -s, t to - t, u to -u and v to -v then we have: x^5 + (r+s+t+u+v)x^4 + (rs+rt+ru+rv+st+su+sv+tu+tv+uv)x^3 + (rst + rsu + rsv + rtu + rtv + ruv + stu + stv + suv + tuv)x^2 + (rstu + rstv + rtuv + stuv + rsuv)x + (rstuv) = 0.
as you've notice the second, fourth and sixth term change in sign therefore we can conclude that x^5 + 9x^4 + 13x^3 - 57x^2 - 86x + 120 = 0 will become x^5 - 9x^4 + 13x^3 + 57x^2 - 86x - 120 = 0.
Since we are looking for |a + b + c + d + e| then it will be |-9 + 13 + 57 - 86 -120| = |-145| = 145.
Can you take 5 minutes and consider if there is a better approach to present your solution?
What do you notice about the coefficients? Is there a pattern?
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@Calvin Lin I think my solution faintly gives the idea to approach this question.
thx
We want to find a polynomial whose roots are negative of the given roots of a given polynomial. To do so we replace x by − x , but since the polynomial is to be monic, we multiply the obtained polynomial by − 1 . Thus the required polynomial is
− ( − x ) 5 + 9 ( − x ) 4 + 1 3 ( − x ) 3 − 5 7 ( − x ) 3 − 8 6 ( − x ) + 1 2 0 ) = f ( x ) ( s a y ) = x 5 − 9 x 4 + 1 3 x 3 + 5 7 x 2 − 8 6 x − 1 2 0 which is same as x 5 + a x 4 + b x 3 + c x 2 + d x + e .
Now ∣ a + b + c + d + e ∣ = ∣ f ( 1 ) − 1 ∣ = 1 4 5 .
First let’s assume f(x)=x5+ax4+bx3+cx2+dx+e and g(x)=x5+9x4+13x3-57x2-86x+120. If r1 is a root of g(x), then g(r1)=r1¬5+9r14+13r13-57r12-86r1+120=0. . . .(1) and f(-r1)=-r15+ar14-br13+cr12-dr1+e. . . . . . (2). If we multiply both sides of (2) by (-1) we get r15-ar14+br13-cr12+dr1-e=0. . . . . (3) Comparing (3) with (1) we get a=-9, b=13, c=57, d=-86, e=-120. Then doing the calculation, mod of(a+b+c+d+e)=mod of (-145)=145
x 5 + 9 x 4 + 1 3 x 3 − 5 7 x 2 − 8 6 x + 1 2 0 = ( x + 5 ) ( x + 4 ) ( x + 3 ) ( x − 1 ) ( x − 2 ) Equating this to zero we get the roots:
r 1 = − 5
r 2 = − 4
r 3 = − 3
r 4 = 1
r 5 = 2
The negative of each roots are the roots of x 5 + a x 4 + b x 3 + c x 2 + d x + e = 0 .
( x − 5 ) ( x − 4 ) ( x − 3 ) ( x + 1 ) ( x + 2 ) = x 5 − 9 x 4 + 1 3 x 3 + 5 7 x 2 − 8 6 x − 1 2 0
∣ a + b + c + d + e ∣ = ∣ − 9 + 1 3 + 5 7 − 8 6 − 1 2 0 ∣ = 1 4 5 .