2017 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 24 Hard

For certain real numbers a, b, and c, the polynomial g(x) = x^3 + ax^2 + x+ 10 has three distinct roots, and each root of g(x) is also a root of the polynomial f(x)= x^4+ x^3+ bx^2+ 100x+c. What is f(1)? (2017 AMC 10A Problem, Question#24)

  • A.

    -9009

  • B.

    -8008

  • C.

    -7007

  • D.

    -6006

  • E.

    -5005

Answer:C

f(x) must have four roots, three of which are roots of g(x). Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of f(x) and g(x) are the same, we know that f(x)=g(x)(x-r)

where r∈C is the fourth root of f(x). Substituting g(x) and expanding, we find that f(x)=(x^3+ax^2+x+10)(x-r)

=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r. Comparing coefficients with f(x), we see that

a-r=1,

1-ar=b,

10 -r=100,

-10 r=c.

(Solution 1. 1 picks up here.)

Let's solve for a, b, c, and r. Since 10 -r= 100, r= -90, so c=(-10)(-90) = 900. Since a -r=1, a=-89, and b=1-ar=-8009. Thus, we know that

f(x) = x^4+x^3-8009x^2+100 x+900.

Taking f(1), we find that

f(1) =1^4+1^3 - 8009(1)^2+ 100(1) + 900,

=1+1- 8009 + 100+ 900,

=(\rm C)-7007.

A faster ending to Solution 1 is as follows. We shall solve for only a and r.

Since 10 - r = 100, r= -90, and since a-r=1, a=-89.Then,

f(1)= (1-r)(1^3 +a·1^2+1+ 10),

= (91)(-77),

=(\rm C)-7007.

We notice that the constant term of f(x)=c and the constant term in g(x)= 10. Because f(x) can be factored as g(x)·(x-r) (where ris the unshared root of f(x), we see that using the constant

Nowwe once agan wile f(x) out in factored fom term, -10\cdot r = c and therefore r=- \dfrac{c}{10}. Now we once again write f(x) out in factored form:

f(x)=g(x) \cdot (x-r)=(x^{3}+ax^{2}+x+10)(x+ \dfrac{c}{10}).

We can expand the expression on the right-hand side to get:

f(x)=x^{4}+(a+ \dfrac{c}{10})x^{3}+(1+ \dfrac{ac}{10})x^{2}+(10+ \dfrac{c}{10})x+c

Now we have

f(x)=x^{4}+(a+ \dfrac{c}{10})x^{3}+(1+ \dfrac{ac}{10})x^{2}+\left( 10+\dfrac{c}{10}\right)x+c=x^{4}+x^{3}+bx^{2}+100x+c.

Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations

10+ \dfrac{c}{10}=100 \Rightarrow c=900a+ \dfrac{c}{10}=1, c=900 ⇒a+90=1⇒a=-89 and finally,

1+ \dfrac{ac}{10}=b=1+ \dfrac{-89 \cdot 900}{10}=b=-8009.

We know that f(1) is the sum of its coefficients, hence 1+ 1+ b + 100 + c. We substitute the values we obtained for b and c into this expression to get f(1)= 1+1+(-8009)+100+900 =(\rm C) -7007.

Let r_1, r_2, and r_3 be the roots of g(x), Let r^4 be the additional root of f(x).Then from Vieta's formulas on the quadratic term of g(x) and the cubic term of f(x), we obtain the following:

r_1+r-2+r_3=-a,

r_1+r_2+r_3 +r_4=-1,

Thus r_4=a-1.

Now applying Vieta's formulas on the constant term of g(x), the linear term of g(x), and the linear term of f(x), we obtain:

r_1r_ 2r_ 3 =-10.

r_1r_ 2+r_2r_ 3+r_3r_1 =1.

r_1r_2r 3+r_2r_3r_ 4+r_3r_4r_1+r_4r_1r_2=-100.

Substituting for r_1 r_2 r_3 in the bottom equation and factoring the remainder of the expression, we obtain:

-10+(r_1 r_ 2+r_2 r_ 3+r_3 r_ 1)r_4=-10+r_4=-100.

It follows that r_4=-90. But r_4 =a-1 so a=-89.

Now we can factor f(x) in terms of g(x) as f(x)=(x-r_4)g(x)= (x+ 90)g(x).

Then f(1)=91g(1)and g(1) = 1^3 - 89·1^2+1+10 = -77.

Hence f(1)= 91\cdot(-77)=(\rm C)- 7007.

Let the roots of g(x) be r_1, r_2, and r_3. Let the roots of f(x) be r_1, r_2, r_3, and r_4. From Vieta's, we

r_1+r_2+r_3 =-a，

r_1+r_2+r_3+r_4=-1．

have: r_4=a-1 The fourth root is a - 1. Since r_1, r_2, and r_3 are common roots, we

f(x) = g(x)(x - (a-1)).

f(1) = g(1)(1 - (a - 1)).

f(1) = (a+ 12)(2 - a).

have: f(1)=-(a+12)(a-2) Let a-2=k ,f(1)=-k(k+14)Nole that -7007 = -1001·(7)= -(7·(11)·(13))·(7)= -91·(77)This gives us a pretty good guess of (\rm C)-7007.

First off, let's get rid of the x^4 term by finding h(x)=f(x)- xg(x).This polynomial consists of the difference of two polynomials with 3 common factors, so it must also have these factors. The polynomial is h(x)=(1-a)x^3+(b-1)x^2+90x +c, and must be equal to (1 - a)g(x).

Equating the coefficients, we get 3 equations. We will tackle the situation one equation at a time, starting the x tems. Looking at the coefficients, we get \dfrac{90}{1-a}=1. 90 =1-a.The solution to the previous is obviously a =-89. We can now find b and c. \dfrac{b-1}{1-a}=a

∴b-1=a(1-a)=-89*90=-8010 and b= -8009. Fmaly \dfrac{c}{1-a}=10,

∴c=10(1-a)= 10*90 = 900 Solving the original problem .f(1)=1+1+b+100+1=102+b+c=102+900-8009= (\rm C)-7007.

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