Question

Show all work to factor x^4 − 17x^2 + 16 completely.

Answer 1

Answer:

$x^{4}-17x^{2} +16$ = (x -1)(x + 1)(x - 4)(x + 4)

Step-by-step explanation:

At first, let us find the first two factors of $x^{4}-17x^{2} +16$

∵ The sign of the last term is positive

∴ The middle signs of the two factors are the same

∵ The sign of the middle term is negative

∴ The middle signs of the two factors are negative

∵ $x^{4}$ = x² × x² ⇒ first terms of the two factors

∵ 16 = -1 × -16 ⇒ second terms of the two factors

∵ x²(-1) + x²(-16) = -x² + -16x² = -17x² ⇒ the value of the middle term

∴ (x² - 1) and (x² - 16) are the factors of $x^{4}-17x^{2} +16$

Now let us factorize each factor

→ The factors of the binomial a² - b² (difference of two squares) are

   (a - b) and (a + b)

∵ x² - 1 is the difference of two squares

∴ Its factors are (x - 1) and (x + 1)

∵ x² - 16 is the difference of two squares

∴ Its factors are (x - 4) and (x + 4)

∵ (x -1), (x + 1), (x - 4), and (x + 4) are the factors of (x² - 1) and (x² - 16)

∵ (x² - 1) and (x² - 16) are the factors of $x^{4}-17x^{2} +16$

∴ (x -1), (x + 1), (x - 4), and (x + 4) are the factors of $x^{4}-17x^{2} +16$

∴  $x^{4}-17x^{2} +16$ = (x -1)(x + 1)(x - 4)(x + 4)