Answer :
Certainly! Let's solve the given polynomial step-by-step:
Given polynomial:
[tex]\[ P(x) = x^4 + 50x^2 + 625 \][/tex]
### (a) Find all zeros of [tex]\( P \)[/tex], real and complex.
To find the zeros of [tex]\( P \)[/tex], we need to solve [tex]\( P(x) = 0 \)[/tex]:
[tex]\[ x^4 + 50x^2 + 625 = 0 \][/tex]
Let's use a substitution to simplify this equation. Let [tex]\( u = x^2 \)[/tex]. Then the polynomial becomes:
[tex]\[ u^2 + 50u + 625 = 0 \][/tex]
Now, we solve this quadratic equation in [tex]\( u \)[/tex]. To solve for [tex]\( u \)[/tex], we can either use factoring, completing the square, or the quadratic formula. In this case, factoring or recognizing patterns might be simpler, but let's proceed with the quadratic formula for clarity:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 50 \)[/tex], and [tex]\( c = 625 \)[/tex].
Substitute in the values:
[tex]\[ u = \frac{-50 \pm \sqrt{50^2 - 4 \cdot 1 \cdot 625}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{-50 \pm \sqrt{2500 - 2500}}{2} \][/tex]
[tex]\[ u = \frac{-50 \pm \sqrt{0}}{2} \][/tex]
[tex]\[ u = \frac{-50}{2} \][/tex]
[tex]\[ u = -25 \][/tex]
Now, recall that [tex]\( u = x^2 \)[/tex]. Therefore,
[tex]\[ x^2 = -25 \][/tex]
To find [tex]\( x \)[/tex], take the square root of both sides:
[tex]\[ x = \pm \sqrt{-25} \][/tex]
[tex]\[ x = \pm 5i \][/tex]
Thus, the zeros of [tex]\( P \)[/tex] are:
[tex]\[ x = 5i, -5i \][/tex]
### (b) Factor [tex]\( P \)[/tex] completely.
Given that the zeros include [tex]\( \pm 5i \)[/tex], we know that [tex]\( x - 5i \)[/tex] and [tex]\( x + 5i \)[/tex] are factors of [tex]\( P \)[/tex]. Because these zeros are roots of a polynomial with real coefficients, their conjugates must also be roots. Thus the polynomial can be factored using these roots.
We know that if [tex]\( x = 5i \)[/tex] and [tex]\( x = -5i \)[/tex] are roots, then:
[tex]\[ (x - 5i)(x + 5i) = x^2 - (5i)^2 = x^2 + 25 \][/tex]
Since these appear as repeated roots (multiplicity 2), the polynomial can be factored as:
[tex]\[ P(x) = (x^2 + 25)^2 \][/tex]
Thus, the complete factorization of [tex]\( P \)[/tex] is:
[tex]\[ P(x) = (x^2 + 25)^2 \][/tex]
### Summary:
(a) The zeros of [tex]\( P \)[/tex], real and complex, are:
[tex]\[ x = 5i, -5i \][/tex]
(b) The polynomial [tex]\( P \)[/tex] factored completely is:
[tex]\[ P(x) = (x^2 + 25)^2 \][/tex]
Given polynomial:
[tex]\[ P(x) = x^4 + 50x^2 + 625 \][/tex]
### (a) Find all zeros of [tex]\( P \)[/tex], real and complex.
To find the zeros of [tex]\( P \)[/tex], we need to solve [tex]\( P(x) = 0 \)[/tex]:
[tex]\[ x^4 + 50x^2 + 625 = 0 \][/tex]
Let's use a substitution to simplify this equation. Let [tex]\( u = x^2 \)[/tex]. Then the polynomial becomes:
[tex]\[ u^2 + 50u + 625 = 0 \][/tex]
Now, we solve this quadratic equation in [tex]\( u \)[/tex]. To solve for [tex]\( u \)[/tex], we can either use factoring, completing the square, or the quadratic formula. In this case, factoring or recognizing patterns might be simpler, but let's proceed with the quadratic formula for clarity:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 50 \)[/tex], and [tex]\( c = 625 \)[/tex].
Substitute in the values:
[tex]\[ u = \frac{-50 \pm \sqrt{50^2 - 4 \cdot 1 \cdot 625}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{-50 \pm \sqrt{2500 - 2500}}{2} \][/tex]
[tex]\[ u = \frac{-50 \pm \sqrt{0}}{2} \][/tex]
[tex]\[ u = \frac{-50}{2} \][/tex]
[tex]\[ u = -25 \][/tex]
Now, recall that [tex]\( u = x^2 \)[/tex]. Therefore,
[tex]\[ x^2 = -25 \][/tex]
To find [tex]\( x \)[/tex], take the square root of both sides:
[tex]\[ x = \pm \sqrt{-25} \][/tex]
[tex]\[ x = \pm 5i \][/tex]
Thus, the zeros of [tex]\( P \)[/tex] are:
[tex]\[ x = 5i, -5i \][/tex]
### (b) Factor [tex]\( P \)[/tex] completely.
Given that the zeros include [tex]\( \pm 5i \)[/tex], we know that [tex]\( x - 5i \)[/tex] and [tex]\( x + 5i \)[/tex] are factors of [tex]\( P \)[/tex]. Because these zeros are roots of a polynomial with real coefficients, their conjugates must also be roots. Thus the polynomial can be factored using these roots.
We know that if [tex]\( x = 5i \)[/tex] and [tex]\( x = -5i \)[/tex] are roots, then:
[tex]\[ (x - 5i)(x + 5i) = x^2 - (5i)^2 = x^2 + 25 \][/tex]
Since these appear as repeated roots (multiplicity 2), the polynomial can be factored as:
[tex]\[ P(x) = (x^2 + 25)^2 \][/tex]
Thus, the complete factorization of [tex]\( P \)[/tex] is:
[tex]\[ P(x) = (x^2 + 25)^2 \][/tex]
### Summary:
(a) The zeros of [tex]\( P \)[/tex], real and complex, are:
[tex]\[ x = 5i, -5i \][/tex]
(b) The polynomial [tex]\( P \)[/tex] factored completely is:
[tex]\[ P(x) = (x^2 + 25)^2 \][/tex]
