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Polynomial function
h is defined as
h(x)=2x^(3)-9x^(2)+cx-6, where
c is a constant. If
2x-3 is a factor of the polynomial, then what is the value of
c ?

Polynomial function $h$ is defined as $h(x)=2 x^{3}-9 x^{2}+c x-6$ , where $c$ is a constant. If $2 x-3$ is a factor of the polynomial, then what is the value of $c$ ?

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Write equations of cosine functions using properties

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Q. Polynomial function

h

is defined as

h(x)=2 x^{3}-9 x^{2}+c x-6

, where

c

is a constant. If

2 x-3

is a factor of the polynomial, then what is the value of

c

?

Factor Theorem Application: Since $2x-3$ is a factor of the polynomial $h(x)$ , we can use the Factor Theorem which states that if $(ax-b)$ is a factor of a polynomial, then the polynomial will equal zero when $x$ equals $\frac{b}{a}$ . In this case, we can find the value of $x$ that makes $2x-3$ equal to zero. $\newline$ $2x - 3 = 0$ $\newline$ $2x = 3$ $\newline$ $x = \frac{3}{2}$
Substitute x Value: Now we substitute $x = \frac{3}{2}$ into the polynomial $h(x)$ and set it equal to zero, because $2x-3$ is a factor of $h(x)$ . $h\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^3 - 9\left(\frac{3}{2}\right)^2 + c\left(\frac{3}{2}\right) - 6 = 0$
Calculate Polynomial: Let's calculate each term separately. $\newline$ First term: $2(\frac{3}{2})^3 = 2 \times \frac{27}{8} = \frac{54}{8} = \frac{27}{4}$ $\newline$ Second term: $-9(\frac{3}{2})^2 = -9 \times \frac{9}{4} = -\frac{81}{4}$ $\newline$ Third term: $c(\frac{3}{2}) = \frac{3c}{2}$ $\newline$ Fourth term: $-6$ $\newline$ Now we combine these terms. $\newline$ $\frac{27}{4} - \frac{81}{4} + \frac{3c}{2} - 6 = 0$
Combine Terms: Combine like terms and simplify the equation. $\newline$ $(\frac{27}{4} - \frac{81}{4}) + \frac{3c}{2} - 6 = 0$ $\newline$ $(-\frac{54}{4}) + \frac{3c}{2} - 6 = 0$ $\newline$ $-\frac{27}{2} + \frac{3c}{2} - 6 = 0$
Combine Fractions: Now we need to combine all terms with fractions and the constant term. $\newline$ To combine $-\frac{27}{2}$ and $-6$ , we need to express $-6$ as a fraction with a denominator of $2$ . $\newline$ $-6 = -\frac{12}{2}$ $\newline$ So the equation becomes: $\newline$ $-\frac{27}{2} + \frac{3c}{2} - \frac{12}{2} = 0$ $\newline$ $\frac{-27 - 12 + 3c}{2} = 0$ $\newline$ $\frac{-39 + 3c}{2} = 0$
Solve for Constant $c$ : To find the value of $c$ , we need to solve for $c$ in the equation $\frac{-39 + 3c}{2} = 0$ . Multiply both sides by $2$ to get rid of the denominator. $2 \times \frac{-39 + 3c}{2} = 2 \times 0$ $-39 + 3c = 0$ Now we solve for $c$ . $3c = 39$ $c = \frac{39}{3}$ $c = 13$

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