Which of the following quartic functions has x=−2 and x=−3 as its only two real zeroes?(1 point) Responses x4−5x3+7x2+5x−6 x to the 4th power minus 5 x cubed plus 7 x squared plus 5 x minus 6 x4+5x3+7x2−5x−6 x to the 4th power plus 5 x cubed plus 7 x squared minus 5 x minus 6 x4−5x3+7x2−5x+6 x to the 4th power minus 5 x cubed plus 7 x squared minus 5 x plus 6 x4+5x3+7x2+5x+6 x to the 4th power plus 5 x cubed plus 7 x squared plus 5 x plus 6
To find which quartic function has \( x = -2 \) and \( x = -3 \) as its only two real zeroes, you can use the factor theorem. According to this theorem, if \( x = -2 \) is a root, then \( (x + 2) \) is a factor of the polynomial. If \( x = -3 \) is a root, then \( (x + 3) \) is also a factor.
Since the polynomial is quartic (degree 4), and these are its only real zeroes, the remaining factors must either be a single factor squared (like \( (x + 2)^2 \) or \( (x + 3)^2 \)) or a pair of complex conjugate factors that do not yield any further real zeroes.
To check each provided polynomial:
1. \( x^4 - 5x^3 + 7x^2 + 5x - 6 \)
2. \( x^4 + 5x^3 + 7x^2 - 5x - 6 \)
3. \( x^4 - 5x^3 + 7x^2 - 5x + 6 \)
4. \( x^4 + 5x^3 + 7x^2 + 5x + 6 \)
Let's substitute \( x = -2 \) and \( x = -3 \) into each polynomial to test if they are indeed roots:
1. For \( x = -2 \): \( (-2)^4 - 5(-2)^3 + 7(-2)^2 + 5(-2) - 6 = 16 - 40 + 28 - 10 - 6 = 0 \)
For \( x = -3 \): \( (-3)^4 - 5(-3)^3 + 7(-3)^2 + 5(-3) - 6 = 81 + 135 + 63 - 15 - 6 \neq 0 \)
2. For \( x = -2 \): \( (-2)^4 + 5(-2)^3 + 7(-2)^2 - 5(-2) - 6 = 16 - 40 + 28 + 10 - 6 = 8 \neq 0 \)
For \( x = -3 \): \( (-3)^4 + 5(-3)^3 + 7(-3)^2 - 5(-3) - 6 = 81 - 135 + 63 + 15 - 6 = 18 \neq 0 \)
3. For \( x = -2 \): \( (-2)^4 - 5(-2)^3 + 7(-2)^2 - 5(-2) + 6 = 16 - 40 + 28 - 10 + 6 = 0 \)
For \( x = -3 \): \( (-3)^4 - 5(-3)^3 + 7(-3)^2 - 5(-3) + 6 = 81 + 135 + 63 - 15 + 6 = 270 \neq 0 \)
4. For \( x = -2 \): \( (-2)^4 + 5(-2)^3 + 7(-2)^2 + 5(-2) + 6 = 16 - 40 + 28 - 10 + 6 = 0 \)
For \( x = -3 \): \( (-3)^4 + 5(-3)^3 + 7(-3)^2 + 5(-3) + 6 = 81 - 135 + 63 - 15 + 6 = 0 \)
Option 4 accurately returns 0 for both \( x = -2 \) and \( x = -3 \), meaning that \( x = -2 \) and \( x = -3 \) are roots of this function, and since it's a quartic, it might contain these roots each to the second power or complex roots to ensure no other real zeroes exist.
Therefore, the function that has \( x = -2 \) and \( x = -3 \) as its only two real zeroes is:
\( x^4 + 5x^3 + 7x^2 + 5x + 6 \)