Question

Which one of the following equations cannot be solved by factoring with integer coefficients? (A) \(12 x^{2}-15 x-63=0\) (B) \(12 x^{2}+46 x-8=0\) (C) \(6 x^{2}-38 x-28=0\) (D) \(8 x^{2}-49 x-68=0\)

Short answer

The equation which cannot be factored with integer coefficients is (B) \(12 x^{2}+46 x-8=0\).

Step-by-step solution

Solving equation (A)

Factorizing the coefficients of (A) \(12 x^{2}-15 x-63=0\) gives that \((4x - 7)(3x + 9) = 0\). Therefore, (A) can be factored with integer coefficients.

Solving equation (B)

Factorizing the coefficients of (B) \(12 x^{2}+46 x-8=0\) fails to produce an integer solution. Therefore, (B) cannot be factored with integer coefficients.

Solving equation (C)

Factorizing the coefficients of (C) \(6 x^{2}-38 x-28=0\) gives that \((2x - 7)(3x - 4) = 0\). Therefore, (C) can be factored with integer coefficients.

Solving equation (D)

Factorizing the coefficients of (D) \(8 x^{2}-49 x-68=0\) gives that \((4x - 17)(2x - 4) = 0\). Therefore, (D) can be factored with integer coefficients.

Key concepts

Integer Coefficients

Integer coefficients in quadratic equations are simply the numbers before the variable terms that are whole numbers. In the context of factoring, these coefficients play a crucial role in determining whether a quadratic equation can be factored easily.

For instance, when we look at the equation mentioned in Step 1, all the coefficients are integers: 12, -15, and -63. To factor this equation, one tries to find two binomials with integer coefficients that multiply together to give the original quadratic. When an equation like (A) can be factored into $\backslash ((4x\; -\; 7)(3x\; +\; 9)\; =\; 0\backslash )$, we observe that the resulting factors also have integer coefficients.

However, sometimes, as shown in Step 2, you can't find such binomial factors with integer coefficients that multiply back to the original quadratic, as with equation (B). This is often the case when the discriminant (the part under the square root in the quadratic formula) is not a perfect square, which suggests that the roots of the equation are not rational numbers, and thus, the equation can't be factored with integers alone.

Quadratic Polynomials

Quadratic polynomials are expressions of the form $\backslash (ax^2\; +\; bx\; +\; c\backslash )$, where $\backslash (a\backslash )$, $\backslash (b\backslash )$, and $\backslash (c\backslash )$ are constants, and $\backslash (a\backslash )$ is not zero. The highest power of the variable, $\backslash (x\backslash )$, is 2, which characterizes these polynomials as quadratic.

In the exercise provided, each of the equations (A), (B), (C), and (D) represents a quadratic polynomial with varying coefficients. We analyze these to find if they can be decomposed into products of binomials with integer coefficients - the essence of factoring them.

Recognizing Quadratic Polynomials

If you're given a polynomial, how can you tell it's quadratic? Check for these features:

• It has three terms: a quadratic term ($\backslash (ax^2\backslash )$), a linear term ($\backslash (bx\backslash )$), and a constant term ($\backslash (c\backslash )$).
• The highest exponent on $\backslash (x\backslash )$ is 2.
• The coefficient $\backslash (a\backslash )$ is not zero because if it were, the polynomial would not be quadratic but linear.

Each equation in the exercise matches these criteria, confirming they are indeed quadratic polynomials.

Solving Quadratic Equations

Solving quadratic equations can be approached in several ways, including factoring, completing the square, graphing, and using the quadratic formula. Whenever possible, factoring is often the quickest method. However, it relies on the polynomial's coefficients and whether they lend themselves to a neat factorization.

In our exercise, we attempted to solve the equations by factoring. When successful, as with (A), (C), and (D), the solutions are found by setting each binomial equal to zero. From (A), $\backslash ((4x\; -\; 7)(3x\; +\; 9)\; =\; 0\backslash )$, we derive $\backslash (x\; =\; \backslash frac\{7\}\{4\}\backslash )$ or $\backslash (x\; =\; -3\backslash )$, as either product being zero makes the equation true.

When Factoring Fails

Sometimes, as with equation (B), no integers can be found that satisfy the necessary factorization. At this juncture, other methods must be employed. The quadratic formula, $\backslash (x\; =\; \backslash frac\{-b\; \backslash pm\; \backslash sqrt\{b^2\; -\; 4ac\}\}\{2a\}\backslash )$, is a reliable catch-all solution that will work when factoring does not. It's important to know this alternative as some quadratic equations, especially those with irrational or complex roots, are unsolvable by simple factorization with integer coefficients.