Answer :
To identify all the roots of the polynomial [tex]\( g(x) = \left(x^2 + 3x - 4\right)\left(x^2 - 4x + 29\right) \)[/tex], we need to solve for [tex]\( x \)[/tex] in each of the two quadratic equations within the polynomial. Let's go through each quadratic equation in detail.
### 1. Solve [tex]\( x^2 + 3x - 4 = 0 \)[/tex]
To solve the quadratic equation [tex]\( x^2 + 3x - 4 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -4 \)[/tex]. Plugging these values into the quadratic formula gives:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 16}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{25}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm 5}{2} \][/tex]
This results in two solutions:
[tex]\[ x = \frac{-3 + 5}{2} = 1 \][/tex]
[tex]\[ x = \frac{-3 - 5}{2} = -4 \][/tex]
Thus, the roots of [tex]\( x^2 + 3x - 4 = 0 \)[/tex] are [tex]\( x = 1 \)[/tex] and [tex]\( x = -4 \)[/tex].
### 2. Solve [tex]\( x^2 - 4x + 29 = 0 \)[/tex]
Similarly, for the second quadratic equation [tex]\( x^2 - 4x + 29 = 0 \)[/tex], we use the quadratic formula with [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 29 \)[/tex]:
[tex]\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 29}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{16 - 116}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{-100}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm 10i}{2} \][/tex]
This results in two complex solutions:
[tex]\[ x = \frac{4 + 10i}{2} = 2 + 5i \][/tex]
[tex]\[ x = \frac{4 - 10i}{2} = 2 - 5i \][/tex]
Thus, the roots of [tex]\( x^2 - 4x + 29 = 0 \)[/tex] are [tex]\( x = 2 + 5i \)[/tex] and [tex]\( x = 2 - 5i \)[/tex].
### Conclusion
Combining all the solutions, the roots of the polynomial [tex]\( g(x) = \left(x^2 + 3x - 4\right)\left(x^2 - 4x + 29\right) \)[/tex] are:
- [tex]\( x = 1 \)[/tex]
- [tex]\( x = -4 \)[/tex]
- [tex]\( x = 2 + 5i \)[/tex]
- [tex]\( x = 2 - 5i \)[/tex]
Therefore, the correct roots, considering the provided options, are:
[tex]\[ \mathbf{1, -4, 2 + 5i, 2 - 5i} \][/tex]
### 1. Solve [tex]\( x^2 + 3x - 4 = 0 \)[/tex]
To solve the quadratic equation [tex]\( x^2 + 3x - 4 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -4 \)[/tex]. Plugging these values into the quadratic formula gives:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 16}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{25}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm 5}{2} \][/tex]
This results in two solutions:
[tex]\[ x = \frac{-3 + 5}{2} = 1 \][/tex]
[tex]\[ x = \frac{-3 - 5}{2} = -4 \][/tex]
Thus, the roots of [tex]\( x^2 + 3x - 4 = 0 \)[/tex] are [tex]\( x = 1 \)[/tex] and [tex]\( x = -4 \)[/tex].
### 2. Solve [tex]\( x^2 - 4x + 29 = 0 \)[/tex]
Similarly, for the second quadratic equation [tex]\( x^2 - 4x + 29 = 0 \)[/tex], we use the quadratic formula with [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 29 \)[/tex]:
[tex]\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 29}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{16 - 116}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{-100}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm 10i}{2} \][/tex]
This results in two complex solutions:
[tex]\[ x = \frac{4 + 10i}{2} = 2 + 5i \][/tex]
[tex]\[ x = \frac{4 - 10i}{2} = 2 - 5i \][/tex]
Thus, the roots of [tex]\( x^2 - 4x + 29 = 0 \)[/tex] are [tex]\( x = 2 + 5i \)[/tex] and [tex]\( x = 2 - 5i \)[/tex].
### Conclusion
Combining all the solutions, the roots of the polynomial [tex]\( g(x) = \left(x^2 + 3x - 4\right)\left(x^2 - 4x + 29\right) \)[/tex] are:
- [tex]\( x = 1 \)[/tex]
- [tex]\( x = -4 \)[/tex]
- [tex]\( x = 2 + 5i \)[/tex]
- [tex]\( x = 2 - 5i \)[/tex]
Therefore, the correct roots, considering the provided options, are:
[tex]\[ \mathbf{1, -4, 2 + 5i, 2 - 5i} \][/tex]