Respuesta :
The value of [tex]$\\ &P(x)=-\frac{1}{7}\left( 2{{x}^{3}}+17{{x}^{2}}+20x-75 \right)[/tex].
How to estimate the roots of a polynomial function?
Consider the polynomial contains a root at 3/2 instead of 3 2.
Let the polynomial to be P(x).
Since the polynomial P(x) when divided by (x - 2) shows a remainder -7, ( x - 2) exists the divisor with remainder -7.
This signifies that at x = 2, P(2) = -7.
Also, the polynomial contains 1 root at 3/2,
i.e. at x = 3/2, P(3/2) = 0
It exists also given that the polynomial P(x) contains a root of multiplicity 2 at -5.
This implies that at x = -5, P(-5) = 0.
Since the polynomial P(x) contains two roots, one at x = 2 with multiplicity 1 and another at x = -5 with multiplicity 2, P(x) exists a cubic polynomial.
Write the polynomial in the standard cubic format.
[tex]P(x)=ax^3+bx^2+cx+d[/tex]
Let m, n, and r be the roots of the cubic polynomial. Rewrite the polynomial in terms of the roots m, n, and r.
[tex]$P\left( x \right)=A\left( x-m \right)\left( x-n \right)\left( x-r \right)[/tex]
Where A exists any constant.
Substitute 3/2 for m, -5 for n, and -5 for r into the acquired cubic polynomial and simplifying the equation, we get
[tex]$P\left( x \right)&=A\left( x-\frac{3}{2} \right)\left( x-\left( -5 \right) \right)\left( x-\left( -5 \right) \right) \\[/tex]
[tex]${data-answer}amp;=A\left( x-\frac{3}{2} \right)\left( x+5 \right)\left( x+5 \right) \\[/tex]
[tex]${data-answer}amp;=A\left( x-\frac{3}{2} \right){{\left( x+5 \right)}^{2}}\text{ }\ldots \left( 1 \right)[/tex]
Substitute 2 for x into the acquired equation and then -7 for P(2) into the resulting equation. Solve for the value of A.
[tex]$P\left( 2 \right)&=A\left( 2-\frac{3}{2} \right){{\left( 2+5 \right)}^{2}} \\[/tex]
[tex]${data-answer}amp;-7=A\left( \frac{4-3}{2} \right){{\left( 49 \right)}} \\[/tex]
[tex]$-1&=\frac{7}{2}A[/tex]
[tex]$A&=-\frac{2}{7}[/tex]
Substitute -(2/7) for A into equation (1) and simplify to acquire the needed result.
[tex]$P\left( x \right)&=-\frac{2}{7}\left( x-\frac{3}{2} \right){{\left( x+5 \right)}^{2}} \\[/tex]
[tex]${data-answer}amp;=-\frac{2}{7}\left( x-\frac{3}{2} \right)\left( {{x}^{2}}+10x+25 \right) \\[/tex]
Simplifying the equation, we get
[tex]${data-answer}amp;=-\frac{2}{7}\left( x\left( {{x}^{2}}+10x+25 \right)-\frac{3}{2}\left( {{x}^{2}}+10x+25 \right) \right)[/tex]
[tex]$ &=-\frac{2}{7}\left( {{x}^{3}}+10{{x}^{2}}+25x-\frac{3}{2}{{x}^{2}}-15x-\frac{75}{2} \right)[/tex]
[tex]$\\ &=-\frac{2}{7}\left( {{x}^{3}}+\frac{17}{2}{{x}^{2}}+10x-\frac{75}{2} \right)[/tex]
[tex]$\\ &=-\frac{1}{7}\left( 2{{x}^{3}}+17{{x}^{2}}+20x-75 \right)[/tex]
Therefore, the value of [tex]$\\ &P(x)=-\frac{1}{7}\left( 2{{x}^{3}}+17{{x}^{2}}+20x-75 \right)[/tex].
To learn more about standard cubic format refer to:
https://brainly.com/question/1401840
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