Soru

Zorluk: ZorFactoring Polynomials

When the polynomial 4x4−37x2+94x^4 - 37x^2 + 9 is factored completely into linear factors of the form ax+bax + b, where aa and bb are integers and a>0a > 0, which of the following expressions represents the sum of these linear factors?

  1. A
    4x4x
  2. B
    6x−86x - 8
  3. 6x6xCevap
  4. D
    6x46x^4
  5. E
    6x−66x - 6

Cevap

The sum of the linear factors is 6x6x.
Factoring the polynomial 4x4−37x2+94x^4 - 37x^2 + 9 by substituting u=x2u = x^2 yields (4u−1)(u−9)(4u - 1)(u - 9), which becomes (4x2−1)(x2−9)(4x^2 - 1)(x^2 - 9). Applying the difference of squares identity to both terms results in the four linear factors (2x−1)(2x - 1), (2x+1)(2x + 1), (x−3)(x - 3), and (x+3)(x + 3). The sum of these factors is 6x6x.

Adım Adım Çözüm

1
Substitute u=x2u = x^2 to rewrite the quartic polynomial as a quadratic expression.
4u2−37u+94u^2 - 37u + 9
This simplifies the polynomial from degree 4 to degree 2, making it easier to factor.
2
Factor the quadratic expression by grouping or finding two numbers that multiply to 3636 and add to −37-37.
(4u−1)(u−9)(4u - 1)(u - 9)
The numbers are −36-36 and −1-1. Rewriting and grouping gives 4u(u−9)−1(u−9)=(4u−1)(u−9)4u(u - 9) - 1(u - 9) = (4u - 1)(u - 9).
3
Substitute x2x^2 back in place of uu and factor the resulting difference of squares binomials.
(2x−1)(2x+1)(x−3)(x+3)(2x - 1)(2x + 1)(x - 3)(x + 3)
Since 4x2−1=(2x)2−124x^2 - 1 = (2x)^2 - 1^2 and x2−9=x2−32x^2 - 9 = x^2 - 3^2, both binomials can be factored completely using the difference of squares identity.
4
Sum the four linear factors.
6x6x
Combining the like terms gives (2x−1)+(2x+1)+(x−3)+(x+3)=(2x+2x+x+x)+(−1+1−3+3)=6x+0=6x(2x - 1) + (2x + 1) + (x - 3) + (x + 3) = (2x + 2x + x + x) + (-1 + 1 - 3 + 3) = 6x + 0 = 6x.

Anahtar Kavram

Factoring quartic polynomials using quadratic substitution and the difference of squares identity.
Bu soruyu puanla