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Find the quadratic term, linear term, constant term, and the values of [tex]a, b,[/tex] and [tex]c[/tex].

[tex]\[
(2x + 5)(x - 1) = -6
\][/tex]


Sagot :

Sure, let's solve the equation and find the quadratic, linear, and constant terms along with the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex].

Given equation:
[tex]\[ (2x + 5)(x - 1) + 6 = 0 \][/tex]

First, we need to expand the product [tex]\((2x + 5)(x - 1)\)[/tex].

1. Expand the product:
[tex]\[ (2x + 5)(x - 1) = 2x \cdot x + 2x \cdot (-1) + 5 \cdot x + 5 \cdot (-1) \][/tex]
Calculate each term:
[tex]\[ = 2x^2 - 2x + 5x - 5 \][/tex]
Combine like terms:
[tex]\[ = 2x^2 + 3x - 5 \][/tex]

2. Add the constant 6 to this result:
[tex]\[ 2x^2 + 3x - 5 + 6 \][/tex]

3. Simplify the equation:
[tex]\[ 2x^2 + 3x + 1 = 0 \][/tex]

So the expanded equation is:
[tex]\[ 2x^2 + 3x + 1 = 0 \][/tex]

From this equation, we can identify the coefficients of the terms:

- The quadratic term [tex]\(a\)[/tex] is the coefficient of [tex]\(x^2\)[/tex]:
[tex]\[ a = 2 \][/tex]

- The linear term [tex]\(b\)[/tex] is the coefficient of [tex]\(x\)[/tex]:
[tex]\[ b = 3 \][/tex]

- The constant term [tex]\(c\)[/tex]:
[tex]\[ c = 1 \][/tex]

Therefore, the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are:
[tex]\[ a = 2, \quad b = 3, \quad c = 1 \][/tex]

These coefficients tell us that the quadratic function in standard form is [tex]\(2x^2 + 3x + 1\)[/tex].
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