To find two trinomials \( f(x) \) and \( g(x) \) such that their sum equals \( 3x - 2 \), we can start by expressing \( f(x) \) and \( g(x) \) in the general trinomial form:
\[ f(x) = ax^2 + bx + c \] \[ g(x) = dx^2 + ex + f \]
We want \( f(x) + g(x) = 3x - 2 \). Thus, we can combine these:
\[ f(x) + g(x) = (a + d)x^2 + (b + e)x + (c + f) \]
This gives us the following system of equations by comparing coefficients with \( 3x - 2 \):
- \( a + d = 0 \) (coefficient of \( x^2 \))
- \( b + e = 3 \) (coefficient of \( x \))
- \( c + f = -2 \) (constant term)
Now, we can choose some values to satisfy these equations.
Let’s choose:
- \( a = 1 \) and \( d = -1 \) (so that \( a + d = 0 \))
- \( b = 2 \) and \( e = 1 \) (so that \( b + e = 3 \))
- \( c = 1 \) and \( f = -3 \) (so that \( c + f = -2 \))
Now we can write the trinomials:
\[ f(x) = 1x^2 + 2x + 1 = x^2 + 2x + 1 \]
\[ g(x) = -1x^2 + 1x - 3 = -x^2 + x - 3 \]
Now, let’s check their sum:
\[ f(x) + g(x) = (x^2 + 2x + 1) + (-x^2 + x - 3) \]
Combining like terms:
\[ = (x^2 - x^2) + (2x + x) + (1 - 3) \] \[ = 0 + 3x - 2 \] \[ = 3x - 2 \]
Thus, the two trinomials \( f(x) = x^2 + 2x + 1 \) and \( g(x) = -x^2 + x - 3 \) satisfy the requirement that their sum equals \( 3x - 2 \).