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 $c$ ? $\newline$ $\square$

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Algebra 1

Simplify exponential expressions using exponent rules

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Q. 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

\newline

\square

Apply Factor Theorem: 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 $\frac{b}{a}$ .
Find x value: Let's find the value of x that makes $2x - 3$ equal to zero. We solve the equation $2x - 3 = 0$ for x. $\newline$ $2x = 3$ $\newline$ $x = \frac{3}{2}$
Substitute $x$ into $h(x)$ : Now we substitute $x = \frac{3}{2}$ into the polynomial $h(x)$ and set it equal to zero, because the Factor Theorem tells us that $h\left(\frac{3}{2}\right)$ should be zero if $2x - 3$ is indeed a factor. $\newline$ $h\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^3 - 9\left(\frac{3}{2}\right)^2 + c\left(\frac{3}{2}\right) - 6 = 0$
Calculate $h\left(\frac{3}{2}\right)$ : We calculate each term separately: $\newline$ $\left(\frac{3}{2}\right)^3 = \frac{27}{8}$ $\newline$ $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ $\newline$ Now we substitute these values into the equation: $\newline$ $2\left(\frac{27}{8}\right) - 9\left(\frac{9}{4}\right) + c\left(\frac{3}{2}\right) - 6 = 0$
Simplify terms: Simplify each term: $\newline$ $2\left(\frac{27}{8}\right) = \frac{27}{4}$ $\newline$ $9\left(\frac{9}{4}\right) = \frac{81}{4}$ $\newline$ $c\left(\frac{3}{2}\right) = \frac{3c}{2}$ $\newline$ Now we have: $\newline$ $\frac{27}{4} - \frac{81}{4} + \frac{3c}{2} - 6 = 0$
Combine like terms: Combine like terms and simplify the equation: $\newline$ $(\frac{27}{4} - \frac{81}{4}) + \frac{3c}{2} - 6 = 0$ $\newline$ $(-\frac{54}{4}) + \frac{3c}{2} - 6 = 0$ $\newline$ $-13.5 + \frac{3c}{2} - 6 = 0$
Combine constant terms: Combine the constant terms: $\newline$ $-13.5 - 6 = -19.5$ $\newline$ So we have: $\newline$ $-19.5 + \frac{3c}{2} = 0$
Solve for c: Now we solve for c: $\newline$ $\frac{3c}{2} = 19.5$ $\newline$ Multiply both sides by $\frac{2}{3}$ to isolate c: $\newline$ $c = (19.5) \times (\frac{2}{3})$ $\newline$ $c = \frac{39}{3}$ $\newline$ $c = 13$