Answer :
To determine the value of [tex]\( b \)[/tex] in the given equation, we need to expand the right-hand side and compare the coefficients of like terms on both sides of the equation.
The equation is:
[tex]\[ x^3 + 6x^2 + 7x - 6 = (x^2 + bx - 2)(x + 3) \][/tex]
Let's start by expanding the right-hand side:
[tex]\[ (x^2 + bx - 2)(x + 3) \][/tex]
To do this, use the distributive property (also known as the FOIL method for binomials):
[tex]\[ (x^2 + bx - 2)(x + 3) = x^2(x + 3) + bx(x + 3) - 2(x + 3) \][/tex]
Now, let's expand each term:
[tex]\[ x^2(x + 3) = x^3 + 3x^2 \][/tex]
[tex]\[ bx(x + 3) = bx^2 + 3bx \][/tex]
[tex]\[ -2(x + 3) = -2x - 6 \][/tex]
Combining all these terms, we get:
[tex]\[ x^3 + 3x^2 + bx^2 + 3bx - 2x - 6 \][/tex]
Now, group the like terms together:
[tex]\[ x^3 + (3 + b)x^2 + (3b - 2)x - 6 \][/tex]
Therefore, the expanded form of the right-hand side is:
[tex]\[ x^3 + (3 + b)x^2 + (3b - 2)x - 6 \][/tex]
Next, we compare the coefficients of the corresponding powers of [tex]\( x \)[/tex] on both sides of the original equation.
The given equation is:
[tex]\[ x^3 + 6x^2 + 7x - 6 = x^3 + (3 + b)x^2 + (3b - 2)x - 6 \][/tex]
By comparing the coefficients of each power of [tex]\( x \)[/tex]:
- The coefficient of [tex]\( x^3 \)[/tex] is 1 on both sides, so that checks out.
- For the [tex]\( x^2 \)[/tex] term: [tex]\( 6 = 3 + b \)[/tex]
- For the [tex]\( x \)[/tex] term: [tex]\( 7 = 3b - 2 \)[/tex]
- The constant term is -6 on both sides, so that checks out.
Let's solve the equations for [tex]\( b \)[/tex].
First, solve the equation for the coefficient of [tex]\( x^2 \)[/tex]:
[tex]\[ 6 = 3 + b \][/tex]
Subtract 3 from both sides:
[tex]\[ 6 - 3 = b \][/tex]
[tex]\[ b = 3 \][/tex]
Now, verify it with the [tex]\( x \)[/tex] term equation to ensure consistency:
[tex]\[ 7 = 3b - 2 \][/tex]
Substitute [tex]\( b = 3 \)[/tex]:
[tex]\[ 7 = 3(3) - 2 \][/tex]
[tex]\[ 7 = 9 - 2 \][/tex]
[tex]\[ 7 = 7 \][/tex]
Thus, both conditions are satisfied with [tex]\( b = 3 \)[/tex].
Therefore, the value of [tex]\( b \)[/tex] is [tex]\( \boxed{3} \)[/tex].
The equation is:
[tex]\[ x^3 + 6x^2 + 7x - 6 = (x^2 + bx - 2)(x + 3) \][/tex]
Let's start by expanding the right-hand side:
[tex]\[ (x^2 + bx - 2)(x + 3) \][/tex]
To do this, use the distributive property (also known as the FOIL method for binomials):
[tex]\[ (x^2 + bx - 2)(x + 3) = x^2(x + 3) + bx(x + 3) - 2(x + 3) \][/tex]
Now, let's expand each term:
[tex]\[ x^2(x + 3) = x^3 + 3x^2 \][/tex]
[tex]\[ bx(x + 3) = bx^2 + 3bx \][/tex]
[tex]\[ -2(x + 3) = -2x - 6 \][/tex]
Combining all these terms, we get:
[tex]\[ x^3 + 3x^2 + bx^2 + 3bx - 2x - 6 \][/tex]
Now, group the like terms together:
[tex]\[ x^3 + (3 + b)x^2 + (3b - 2)x - 6 \][/tex]
Therefore, the expanded form of the right-hand side is:
[tex]\[ x^3 + (3 + b)x^2 + (3b - 2)x - 6 \][/tex]
Next, we compare the coefficients of the corresponding powers of [tex]\( x \)[/tex] on both sides of the original equation.
The given equation is:
[tex]\[ x^3 + 6x^2 + 7x - 6 = x^3 + (3 + b)x^2 + (3b - 2)x - 6 \][/tex]
By comparing the coefficients of each power of [tex]\( x \)[/tex]:
- The coefficient of [tex]\( x^3 \)[/tex] is 1 on both sides, so that checks out.
- For the [tex]\( x^2 \)[/tex] term: [tex]\( 6 = 3 + b \)[/tex]
- For the [tex]\( x \)[/tex] term: [tex]\( 7 = 3b - 2 \)[/tex]
- The constant term is -6 on both sides, so that checks out.
Let's solve the equations for [tex]\( b \)[/tex].
First, solve the equation for the coefficient of [tex]\( x^2 \)[/tex]:
[tex]\[ 6 = 3 + b \][/tex]
Subtract 3 from both sides:
[tex]\[ 6 - 3 = b \][/tex]
[tex]\[ b = 3 \][/tex]
Now, verify it with the [tex]\( x \)[/tex] term equation to ensure consistency:
[tex]\[ 7 = 3b - 2 \][/tex]
Substitute [tex]\( b = 3 \)[/tex]:
[tex]\[ 7 = 3(3) - 2 \][/tex]
[tex]\[ 7 = 9 - 2 \][/tex]
[tex]\[ 7 = 7 \][/tex]
Thus, both conditions are satisfied with [tex]\( b = 3 \)[/tex].
Therefore, the value of [tex]\( b \)[/tex] is [tex]\( \boxed{3} \)[/tex].