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A polynomial function qq with real coefficients satisfies the equation q(x)+q(6−x)=8q(x) + q(6-x) = 8 for all real numbers xx. In the xyxy-plane, the graph of y=q(x)y = q(x) has an xx-intercept at (5,0)(5, 0). What is the remainder when q(x)q(x) is divided by x−1x-1?

  1. 88Cevap
  2. B
    −8-8
  3. C
    00
  4. D
    22

Cevap

The correct answer is 88, which is the remainder when the polynomial is divided by x−1x-1.
To find the remainder when the polynomial q(x)q(x) is divided by x−1x-1, we use the Remainder Theorem, which states that this remainder is equal to q(1)q(1). Since the graph of y=q(x)y = q(x) has an xx-intercept at (5,0)(5, 0), we know that q(5)=0q(5) = 0. Substituting x=5x = 5 into the given equation q(x)+q(6−x)=8q(x) + q(6-x) = 8 yields q(5)+q(6−5)=8q(5) + q(6-5) = 8, which simplifies to q(5)+q(1)=8q(5) + q(1) = 8. Substituting q(5)=0q(5) = 0 into this equation gives 0+q(1)=80 + q(1) = 8, so q(1)=8q(1) = 8. Geometrically, the equation q(x)+q(6−x)2=4\frac{q(x) + q(6-x)}{2} = 4 shows that the graph of qq is symmetric under a point reflection about (3,4)(3, 4). Reflecting the point (5,0)(5, 0) about (3,4)(3, 4) gives the point (1,8)(1, 8), verifying that q(1)=8q(1) = 8.

Adım Adım Çözüm

1
Translate the given graphical intercept into function notation.
q(5)=0q(5) = 0
An xx-intercept at (5,0)(5, 0) means that when the input xx is 55, the output of the function is 00.
2
Apply the Remainder Theorem to the divisor x−1x-1.
The remainder is equal to q(1)q(1).
The Remainder Theorem states that when a polynomial q(x)q(x) is divided by x−ax-a, the remainder is q(a)q(a).
3
Substitute x=5x = 5 into the given functional equation q(x)+q(6−x)=8q(x) + q(6-x) = 8.
q(5)+q(6−5)=8  ⟹  q(5)+q(1)=8q(5) + q(6-5) = 8 \implies q(5) + q(1) = 8
Substituting x=5x = 5 allows us to relate the known value q(5)q(5) directly to the target value q(1)q(1).
4
Solve for q(1)q(1) using the value of q(5)q(5).
0+q(1)=8  ⟹  q(1)=80 + q(1) = 8 \implies q(1) = 8
Substituting q(5)=0q(5) = 0 into the equation yields the value of q(1)q(1), which is the remainder.

Anahtar Kavram

Remainder Theorem and Polynomial Graphs

Alternatif Yöntem

Use the point symmetry of the graph. The equation q(x)+q(6−x)=8q(x) + q(6-x) = 8 can be rewritten as q(x)+q(6−x)2=4\frac{q(x) + q(6-x)}{2} = 4. This indicates that the graph of y=q(x)y = q(x) has point symmetry about the point (3,4)(3, 4). Since the graph has an xx-intercept at (5,0)(5, 0), point symmetry dictates that the reflected point across (3,4)(3, 4) must also lie on the graph. The midpoint of (5,0)(5, 0) and the symmetric point (x0,y0)(x_0, y_0) is (3,4)(3, 4). Solving 5+x02=3\frac{5+x_0}{2} = 3 gives x0=1x_0 = 1, and 0+y02=4\frac{0+y_0}{2} = 4 gives y0=8y_0 = 8. Thus, (1,8)(1, 8) lies on the graph, meaning q(1)=8q(1) = 8. By the Remainder Theorem, the remainder when q(x)q(x) is divided by x−1x-1 is q(1)=8q(1) = 8.
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Polynomial Factors and Graphs

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