Answer :
To solve the polynomial equation [tex]\(x^5 - 17x^4 + 70x^3 = 0\)[/tex] algebraically, we can use factoring to find the solutions for [tex]\(x\)[/tex].
Step 1: Factor out the greatest common factor.
The equation is [tex]\(x^5 - 17x^4 + 70x^3 = 0\)[/tex]. We notice that each term has at least [tex]\(x^3\)[/tex] as a factor. We can factor out [tex]\(x^3\)[/tex]:
[tex]\[
x^3(x^2 - 17x + 70) = 0
\][/tex]
Step 2: Solve for [tex]\(x\)[/tex] in each factor.
Now, we have a product of factors equal to zero. According to the zero-product property, if the product of factors is zero, then at least one of the factors must be zero.
1. First factor: [tex]\(x^3 = 0\)[/tex]
To solve [tex]\(x^3 = 0\)[/tex], apply the cube root to both sides:
[tex]\[
x = 0
\][/tex]
So, one solution is [tex]\(x = 0\)[/tex].
2. Second factor: [tex]\(x^2 - 17x + 70 = 0\)[/tex]
This is a quadratic equation. We can solve it using the quadratic formula:
The quadratic formula is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For our quadratic equation [tex]\(x^2 - 17x + 70 = 0\)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = -17\)[/tex]
- [tex]\(c = 70\)[/tex]
Calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[
b^2 - 4ac = (-17)^2 - 4 \cdot 1 \cdot 70 = 289 - 280 = 9
\][/tex]
Now, substitute the values into the quadratic formula:
[tex]\[
x = \frac{-(-17) \pm \sqrt{9}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{17 \pm 3}{2}
\][/tex]
Calculate the two solutions:
[tex]\[
x = \frac{17 + 3}{2} = \frac{20}{2} = 10
\][/tex]
[tex]\[
x = \frac{17 - 3}{2} = \frac{14}{2} = 7
\][/tex]
Therefore, the solutions for [tex]\(x\)[/tex] are [tex]\(x = 0\)[/tex], [tex]\(x = 10\)[/tex], and [tex]\(x = 7\)[/tex].
Final Solution:
The values of [tex]\(x\)[/tex] that satisfy the equation [tex]\(x^5 - 17x^4 + 70x^3 = 0\)[/tex] are [tex]\(x = 0\)[/tex], [tex]\(x = 10\)[/tex], and [tex]\(x = 7\)[/tex].
Step 1: Factor out the greatest common factor.
The equation is [tex]\(x^5 - 17x^4 + 70x^3 = 0\)[/tex]. We notice that each term has at least [tex]\(x^3\)[/tex] as a factor. We can factor out [tex]\(x^3\)[/tex]:
[tex]\[
x^3(x^2 - 17x + 70) = 0
\][/tex]
Step 2: Solve for [tex]\(x\)[/tex] in each factor.
Now, we have a product of factors equal to zero. According to the zero-product property, if the product of factors is zero, then at least one of the factors must be zero.
1. First factor: [tex]\(x^3 = 0\)[/tex]
To solve [tex]\(x^3 = 0\)[/tex], apply the cube root to both sides:
[tex]\[
x = 0
\][/tex]
So, one solution is [tex]\(x = 0\)[/tex].
2. Second factor: [tex]\(x^2 - 17x + 70 = 0\)[/tex]
This is a quadratic equation. We can solve it using the quadratic formula:
The quadratic formula is:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For our quadratic equation [tex]\(x^2 - 17x + 70 = 0\)[/tex], the coefficients are:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = -17\)[/tex]
- [tex]\(c = 70\)[/tex]
Calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[
b^2 - 4ac = (-17)^2 - 4 \cdot 1 \cdot 70 = 289 - 280 = 9
\][/tex]
Now, substitute the values into the quadratic formula:
[tex]\[
x = \frac{-(-17) \pm \sqrt{9}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{17 \pm 3}{2}
\][/tex]
Calculate the two solutions:
[tex]\[
x = \frac{17 + 3}{2} = \frac{20}{2} = 10
\][/tex]
[tex]\[
x = \frac{17 - 3}{2} = \frac{14}{2} = 7
\][/tex]
Therefore, the solutions for [tex]\(x\)[/tex] are [tex]\(x = 0\)[/tex], [tex]\(x = 10\)[/tex], and [tex]\(x = 7\)[/tex].
Final Solution:
The values of [tex]\(x\)[/tex] that satisfy the equation [tex]\(x^5 - 17x^4 + 70x^3 = 0\)[/tex] are [tex]\(x = 0\)[/tex], [tex]\(x = 10\)[/tex], and [tex]\(x = 7\)[/tex].
