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Karo-lina-s [1.5K]
3 years ago
12

How many cubic (i.e., third-degree) polynomials $f(x)$ are there such that $f(x)$ has positive integer coefficients and $f(1)=9$

? (note: all coefficients must be positive---coefficients are not permitted to be 0, so for example $f(x) = x^3 + 8$ is not a valid polynomial.)?
Mathematics
1 answer:
Firdavs [7]3 years ago
6 0
Hello,
there are 56 polynomials : a*x^3+b*x^2+c*x+d=0
num            a  b  c  d
 1             1  1  1  6
 2             1  1  2  5
 3             1  1  3  4
 4             1  1  4  3
 5             1  1  5  2
 6             1  1  6  1
 7             1  2  1  5
 8             1  2  2  4
 9             1  2  3  3
 10            1  2  4  2
 11            1  2  5  1
 12            1  3  1  4
 13            1  3  2  3
 14            1  3  3  2
 15            1  3  4  1
 16            1  4  1  3
 17            1  4  2  2
 18            1  4  3  1
 19            1  5  1  2
 20            1  5  2  1
 21            1  6  1  1
 22            2  1  1  5
 23            2  1  2  4
 24            2  1  3  3
 25            2  1  4  2
 26            2  1  5  1
 27            2  2  1  4
 28            2  2  2  3
 29            2  2  3  2
 30            2  2  4  1
 31            2  3  1  3
 32            2  3  2  2
 33            2  3  3  1
 34            2  4  1  2
 35            2  4  2  1
 36            2  5  1  1
 37            3  1  1  4
 38            3  1  2  3
 39            3  1  3  2
 40            3  1  4  1
 41            3  2  1  3
 42            3  2  2  2
 43            3  2  3  1
 44            3  3  1  2
 45            3  3  2  1
 46            3  4  1  1
 47            4  1  1  3
 48            4  1  2  2
 49            4  1  3  1
 50            4  2  1  2
 51            4  2  2  1
 52            4  3  1  1
 53            5  1  1  2
 54            5  1  2  1
 55            5  2  1  1
 56            6  1  1  1

DIM a AS INTEGER, b AS INTEGER, c AS INTEGER, d AS INTEGER, k AS INTEGER
OPEN "c:\nosdevoirs\polynome.sol" FOR OUTPUT AS #1
k = 0
FOR a = 1 TO 6
    FOR b = 1 TO 6
        FOR c = 1 TO 6
            FOR d = 1 TO 6
                IF a + b + c + d = 9 THEN
                    k = k + 1
                    PRINT k, a; b; c; d
                    PRINT #1, k, a; b; c; d

                END IF
            NEXT d
        NEXT c

    NEXT b
NEXT a
CLOSE #1
END



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The greatest possible number of club members is 7

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Given that, local readers’ club has a set of 49 hardback books and a set of 21 paperbacks

Each set can be divided equally among the club members

To find the greatest possible number of club members, we have to find the greatest common factor of 49 and 21

The greatest number that is a factor of two (or more) other numbers.

When we find all the factors of two or more numbers, and some factors are the same ("common"), then the largest of those common factors is the Greatest Common Factor.

<em><u>Greatest common factor of 49 and 21:</u></em>

The factors of 21 are: 1, 3, 7, 21

The factors of 49 are: 1, 7, 49

Then the greatest common factor is 7

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Write out the equation

\begin{gathered} y=\frac{1}{3}x-3 \\ 5x-3y=-3 \end{gathered}

To solve this system of equation, the aprropriate method will be substitution

Substitute the expression for y into the second equation for y, we have

5x-3(\frac{1}{3}x-3)=-3

Expand the paranthesis we have

\begin{gathered} 5x-x+9=-3 \\ \text{Simplify},\text{ by subtracting 9 from both sides } \\ 4x+9-9=-3-9 \\ 4x=-12 \end{gathered}

Divide both sides by 4, we have

\begin{gathered} \frac{4x}{4}=-\frac{12}{4} \\ \text{Then} \\ x=-3 \end{gathered}

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\begin{gathered} y=\frac{1}{3}x-3 \\ \text{ recall x=}-3 \\ y=\frac{1}{3}(-3)-3 \\ y=-1-4 \\ \text{Hence } \\ y=-4 \end{gathered}

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