Q. \(\,x^2-3x+2\,\)
Answer
We factor the quadratic.
\[
x^2-3x+2=(x-1)(x-2)
\]
So the solutions come from each factor being zero.
\[
x-1=0 \Rightarrow x=1,\quad x-2=0 \Rightarrow x=2
\]
Final answer: \(\,x=1\) or \(x=2\).
Detailed Explanation
We are asked to simplify or factor the expression \(x^2 – 3x + 2\). A common goal for a quadratic like this is to factor it into the form \((x-a)(x-b)\).
Step 1: Identify the coefficients.
The expression is:
\[x^2 – 3x + 2\]
So the coefficients are:
\[a = 1,\quad b = -3,\quad c = 2\]
Step 2: Find two numbers that multiply to \(c\) and add to \(b\).
We need numbers \(m\) and \(n\) such that:
\[m \cdot n = 2\]
\[m + n = -3\]
Check factor pairs of \(2\): \(1 \cdot 2 = 2\) and \((-1) \cdot (-2) = 2\).
Now see which pair adds to \(-3\): the numbers \(-1\) and \(-2\) add to \(-3\).
So we choose:
\[m = -1,\quad n = -2\]
Step 3: Write the factored form.
Using \(m\) and \(n\), the quadratic factors as:
\[x^2 – 3x + 2 = (x + m)(x + n)\]
Substitute \(m = -1\) and \(n = -2\):
\[x^2 – 3x + 2 = (x – 1)(x – 2)\]
Final Answer:
\[x^2 – 3x + 2 = (x – 1)(x – 2)\]
Graph
Algebra FAQ
Factor \(x^2-3x+2\).
Find the zeros of \(x^2-3x+2=0\).
Solve \(x^2-3x+2>0\).
Solve \(x^2-3x+2<0\).
Compute the vertex and minimum value of \(x^2-3x+2\).
Determine the discriminant of \(x^2-3x+2\).
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