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Answer :
Sure! Let's find the real zeros for each of the given polynomials step-by-step.
23. [tex]\( f(x) = x^4 + 2x^2 - 15 \)[/tex]
To find the real zeros of this polynomial, we set the equation equal to zero:
[tex]\[ x^4 + 2x^2 - 15 = 0 \][/tex]
1. Let's make a substitution to simplify the equation. Set [tex]\( y = x^2 \)[/tex]. Then [tex]\( x^4 = y^2 \)[/tex], and the equation becomes:
[tex]\[ y^2 + 2y - 15 = 0 \][/tex]
2. This is a quadratic equation in terms of [tex]\( y \)[/tex]. We can solve it using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -15 \)[/tex].
3. Calculate the discriminant:
[tex]\[ b^2 - 4ac = 2^2 - 4(1)(-15) = 4 + 60 = 64 \][/tex]
4. Substitute into the quadratic formula:
[tex]\[ y = \frac{-2 \pm \sqrt{64}}{2} = \frac{-2 \pm 8}{2} \][/tex]
5. This gives two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{6}{2} = 3 \quad \text{and} \quad y = \frac{-10}{2} = -5 \][/tex]
6. Recall that [tex]\( y = x^2 \)[/tex], so:
\- For [tex]\( y = 3 \)[/tex], [tex]\( x^2 = 3 \)[/tex] giving [tex]\( x = \pm \sqrt{3} \)[/tex].
\- For [tex]\( y = -5 \)[/tex], [tex]\( x^2 = -5 \)[/tex] which gives complex solutions, not real zeros.
Therefore, the real zeros are [tex]\( x = \pm \sqrt{3} \)[/tex].
25. [tex]\( f(x) = x^3 - 7x^2 + x - 7 \)[/tex]
Set the polynomial equal to zero to find its real zeros:
[tex]\[ x^3 - 7x^2 + x - 7 = 0 \][/tex]
1. Start by using the Rational Root Theorem to test possible rational roots, which could be factors of the constant term (-7).
2. You can test values like [tex]\( x = \pm 1, \pm 7 \)[/tex].
3. By substituting [tex]\( x = 7 \)[/tex] into the polynomial, we find it satisfies the equation:
[tex]\[ 7^3 - 7 \times 7^2 + 7 - 7 = 343 - 343 + 7 - 7 = 0 \][/tex]
So, [tex]\( x = 7 \)[/tex] is a real zero of the polynomial.
In conclusion, the real zero for this polynomial is [tex]\( x = 7 \)[/tex].
For the remaining factors, which result in complex solutions, we focus only on real solutions.
23. [tex]\( f(x) = x^4 + 2x^2 - 15 \)[/tex]
To find the real zeros of this polynomial, we set the equation equal to zero:
[tex]\[ x^4 + 2x^2 - 15 = 0 \][/tex]
1. Let's make a substitution to simplify the equation. Set [tex]\( y = x^2 \)[/tex]. Then [tex]\( x^4 = y^2 \)[/tex], and the equation becomes:
[tex]\[ y^2 + 2y - 15 = 0 \][/tex]
2. This is a quadratic equation in terms of [tex]\( y \)[/tex]. We can solve it using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -15 \)[/tex].
3. Calculate the discriminant:
[tex]\[ b^2 - 4ac = 2^2 - 4(1)(-15) = 4 + 60 = 64 \][/tex]
4. Substitute into the quadratic formula:
[tex]\[ y = \frac{-2 \pm \sqrt{64}}{2} = \frac{-2 \pm 8}{2} \][/tex]
5. This gives two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{6}{2} = 3 \quad \text{and} \quad y = \frac{-10}{2} = -5 \][/tex]
6. Recall that [tex]\( y = x^2 \)[/tex], so:
\- For [tex]\( y = 3 \)[/tex], [tex]\( x^2 = 3 \)[/tex] giving [tex]\( x = \pm \sqrt{3} \)[/tex].
\- For [tex]\( y = -5 \)[/tex], [tex]\( x^2 = -5 \)[/tex] which gives complex solutions, not real zeros.
Therefore, the real zeros are [tex]\( x = \pm \sqrt{3} \)[/tex].
25. [tex]\( f(x) = x^3 - 7x^2 + x - 7 \)[/tex]
Set the polynomial equal to zero to find its real zeros:
[tex]\[ x^3 - 7x^2 + x - 7 = 0 \][/tex]
1. Start by using the Rational Root Theorem to test possible rational roots, which could be factors of the constant term (-7).
2. You can test values like [tex]\( x = \pm 1, \pm 7 \)[/tex].
3. By substituting [tex]\( x = 7 \)[/tex] into the polynomial, we find it satisfies the equation:
[tex]\[ 7^3 - 7 \times 7^2 + 7 - 7 = 343 - 343 + 7 - 7 = 0 \][/tex]
So, [tex]\( x = 7 \)[/tex] is a real zero of the polynomial.
In conclusion, the real zero for this polynomial is [tex]\( x = 7 \)[/tex].
For the remaining factors, which result in complex solutions, we focus only on real solutions.
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