If the product of $t^{2}+kt-5$ and $2t-3$ is $2t^{3}-11t^{2}+2t+15$ , then what is the value of $k$ ? $\newline$ A. $-5$ $\newline$ B. $-4$ $\newline$ C. $2$ $\newline$ D. $3$

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Q. If the product of

t^{2}+kt-5

and

2t-3

2t^{3}-11t^{2}+2t+15

, then what is the value of

k

\newline

-5

\newline

-4

\newline

2

\newline

3

Set up equation: Set up the equation for the product of the two polynomials. $\newline$ We are given that $(t^2 + kt - 5) \times (2t - 3) = 2t^3 - 11t^2 + 2t + 15$ .
Distribute terms: Distribute $(2t - 3)$ across each term in $(t^2 + kt - 5)$ . This means we will multiply $2t$ by each term in the first polynomial and then $-3$ by each term in the first polynomial. $(2t \cdot t^2) + (2t \cdot kt) + (2t \cdot -5) + (-3 \cdot t^2) + (-3 \cdot kt) + (-3 \cdot -5)$
Simplify expression: Simplify the expression by combining like terms. $\newline$ $2t^3 + 2kt^2 - 10t - 3t^2 - 3kt + 15$ $\newline$ Now we combine the $t^2$ terms and the $t$ terms. $\newline$ $2t^3 + (2k - 3)t^2 + (-10 - 3k)t + 15$
Compare coefficients: Compare the coefficients of the simplified expression with the given polynomial. $\newline$ We have the expression $2t^3 + (2k - 3)t^2 + (-10 - 3k)t + 15$ and it needs to be equal to $2t^3 - 11t^2 + 2t + 15$ . $\newline$ By comparing coefficients, we get: $\newline$ For $t^3$ : $2 = 2$ (which is already true) $\newline$ For $t^2$ : $2k - 3 = -11$ $\newline$ For $t$ : $-10 - 3k = 2$ $\newline$ For the constant term: $15 = 15$ (which is already true)
Solve for $k$ : Solve the equation $2k - 3 = -11$ for $k$ .
$2k = -11 + 3$
$2k = -8$
$k = -8 / 2$
$k = -4$
Verify solution: Verify the solution by substituting $k = -4$ into the other equation. $\newline$ $-10 - 3(-4) = 2$ $\newline$ $-10 + 12 = 2$ $\newline$ $2 = 2$ $\newline$ This confirms that $k = -4$ is the correct solution.