Solve for x. Enter the solutions from least to greatest. {:[(-5x+4)(x-3)=0],[" lesser "x=◻],[" greater "x=◻]:}

Setting up the equation: To find the roots of the polynomial, we need to set each factor equal to zero and solve for x. First, let's consider the factor (-5x + 4). (-5x + 4) = 0 To solve for x, we add 5x to both sides and then divide by -5. -5x + 4 + 5x = 0 + 5x 4 = 5x x = 4 / 5Solving for x in the first factor: Now, let's consider the second factor (x - 3). (x - 3) = 0 To solve for x, we add 3 to both sides. x - 3 + 3 = 0 + 3 x = 3Solving for x in the second factor: We have found two solutions for x: 4/5 and 3. To enter the solutions from least to greatest, we compare the two values. Since 4/5 is less than 3, we list 4/5 as the ＂lesser＂ x and 3 as the ＂greater＂ x.

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Solve for
x.
Enter the solutions from least to greatest.

{:[(-5x+4)(x-3)=0],[" lesser "x=◻],[" greater "x=◻]:}

Solve for $x$ . $\newline$ Enter the solutions from least to greatest. $\newline$ $\begin{array}{l} (-5 x+4)(x-3)=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}$

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Math Problems

Algebra 2

Find the roots of factored polynomials

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Q. Solve for

x

\newline

Enter the solutions from least to greatest.

\newline

\begin{array}{l} (-5 x+4)(x-3)=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}

Setting up the equation: To find the roots of the polynomial, we need to set each factor equal to zero and solve for x. $\newline$ First, let's consider the factor $(-5x + 4)$ . $\newline$ $(-5x + 4) = 0$ $\newline$ To solve for x, we add $5x$ to both sides and then divide by $-5$ . $\newline$ $-5x + 4 + 5x = 0 + 5x$ $\newline$ $4 = 5x$ $\newline$ $x = \frac{4}{5}$
Solving for x in the first factor: Now, let's consider the second factor $(x - 3)$ . $\newline$ $(x - 3) = 0$ $\newline$ To solve for x, we add $3$ to both sides. $\newline$ $x - 3 + 3 = 0 + 3$ $\newline$ $x = 3$
Solving for x in the second factor: We have found two solutions for x: $\frac{4}{5}$ and $3$ . To enter the solutions from least to greatest, we compare the two values. $\newline$ Since $\frac{4}{5}$ is less than $3$ , we list $\frac{4}{5}$ as the ＂lesser＂ x and $3$ as the ＂greater＂ x.