x, squared, minus, 12, x, plus, 27
x
2
−12x+27
11 answers
To factor the quadratic expression \( x^2 - 12x + 27 \), we can look for two numbers that multiply to \( 27 \) (the constant term) and add to \( -12 \) (the coefficient of the \( x \) term).
The two numbers that meet these criteria are \( -3 \) and \( -9 \) since:
Therefore, we can write the factorization as:
\[ x^2 - 12x + 27 = (x - 3)(x - 9) \]
So, the factored form of \( x^2 - 12x + 27 \) is:
\[ (x - 3)(x - 9) \]
To factor the quadratic expression \( x^2 + 4x - 21 \), we need to find two numbers that multiply to \( -21 \) (the constant term) and add up to \( 4 \) (the coefficient of the \( x \) term).
The two numbers that meet these criteria are \( 7 \) and \( -3 \) since:
Using these numbers, we can factor the expression as follows:
\[ x^2 + 4x - 21 = (x + 7)(x - 3) \]
So, the factored form of \( x^2 + 4x - 21 \) is:
\[ (x + 7)(x - 3) \]
To factor the trinomial \( 5x^2 + 7x + 2 \), we can use the method of finding two numbers that multiply to the product of the first coefficient (\( 5 \)) and the last term (\( 2 \)), which is \( 5 \times 2 = 10 \), and add up to the middle coefficient (\( 7 \)).
The two numbers that satisfy these conditions are \( 5 \) and \( 2 \):
Next, we rewrite the middle term \( 7x \) using these numbers:
\[ 5x^2 + 5x + 2x + 2 \]
Now, we can group the terms:
\[ (5x^2 + 5x) + (2x + 2) \]
Factoring out the common factors in each group, we get:
\[ 5x(x + 1) + 2(x + 1) \]
Now, we can factor out the common binomial factor \( (x + 1) \):
\[ (5x + 2)(x + 1) \]
Thus, the factored form of \( 5x^2 + 7x + 2 \) is:
\[ (5x + 2)(x + 1) \]
To factor the trinomial \( 5x^2 + 32x + 12 \), we need to find two numbers that multiply to \( 5 \times 12 = 60 \) (the product of the coefficient of \( x^2 \) and the constant term) and add up to \( 32 \) (the coefficient of the \( x \) term).
The two numbers that satisfy these conditions are \( 30 \) and \( 2 \):
Now, we can rewrite the middle term \( 32x \) using these two numbers:
\[ 5x^2 + 30x + 2x + 12 \]
Next, we group the terms:
\[ (5x^2 + 30x) + (2x + 12) \]
Now, we can factor out the common factors in each group:
\[ 5x(x + 6) + 2(x + 6) \]
This gives us:
\[ (5x + 2)(x + 6) \]
Thus, the factored form of \( 5x^2 + 32x + 12 \) is:
\[ (5x + 2)(x + 6) \]
To solve the equation
\[ 3(5x + 2)(x^2 - 81) = 0 \]
we can set each factor equal to zero.
\[ 3 = 0 \]
This does not yield any solutions, as \(3\) is a non-zero constant.
\[ 5x + 2 = 0 \]
Solving for \(x\):
\[ 5x = -2 \implies x = -\frac{2}{5} \]
\[ x^2 - 81 = 0 \]
This is a difference of squares, which can be factored as:
\[ (x - 9)(x + 9) = 0 \]
Setting each of these factors equal to zero gives us:
\[ x - 9 = 0 \implies x = 9 \] \[ x + 9 = 0 \implies x = -9 \]
Combining all the solutions, we have:
\[ x = -\frac{2}{5}, \quad x = 9, \quad x = -9 \]
Thus, the complete solution set for the equation \( 3(5x + 2)(x^2 - 81) = 0 \) is:
\[ x = -\frac{2}{5}, \quad x = 9, \quad x = -9 \]
To solve the equation
\[ -2x(x + 8)(x^2 - 9) = 0, \]
we can set each factor equal to zero:
\[ -2x = 0 \]
This gives:
\[ x = 0. \]
\[ x + 8 = 0 \]
This gives:
\[ x = -8. \]
\[ x^2 - 9 = 0. \]
This can be factored as:
\[ (x - 3)(x + 3) = 0. \]
Setting each of these factors equal to zero gives:
\[ x - 3 = 0 \implies x = 3, \] \[ x + 3 = 0 \implies x = -3. \]
Now, combining all the solutions, we have:
\[ x = 0, \quad x = -8, \quad x = 3, \quad x = -3. \]
Thus, the complete solution set for the equation
\[ -2x(x + 8)(x^2 - 9) = 0 \]
is:
\[ x = 0, \quad x = -8, \quad x = 3, \quad x = -3. \]
