Question

1. identify the factors and zeros of the following quadratic equation:

select 4 choices:
$7x^2 + x - 4 = 2$
$\square$ $(x + 1)$
$\square$ $x = 1$
$\square$ $(7x - 6)$
$\square$ $x = \frac{6}{7}$
$\square$ $x = 4$
$\square$ $x = -\frac{6}{7}$
$\square$ $(x - 4)$
$\square$ $x = -1$
$\square$ $(7x + 6)$
clear all

Explanation:

Step1: Rearrange the equation

First, we need to set the quadratic equation to standard form \(ax^2 + bx + c = 0\). Starting with \(7x^2 + x - 4 = 2\), we subtract 2 from both sides:
\[ 7x^2 + x - 4 - 2 = 0 \]
\[ 7x^2 + x - 6 = 0 \]

Step2: Factor the quadratic

We factor the quadratic \(7x^2 + x - 6\). We need two numbers that multiply to \(7\times(-6)= -42\) and add up to 1. The numbers are 7 and -6 (since \(7\times(-6)= -42\) and \(7 + (-6)=1\)). We rewrite the middle term using these numbers:
\[ 7x^2 + 7x - 6x - 6 = 0 \]
Group the terms:
\[ (7x^2 + 7x) + (-6x - 6) = 0 \]
Factor out the greatest common factor from each group:
\[ 7x(x + 1) - 6(x + 1) = 0 \]
Now, factor out the common binomial factor \((x + 1)\):
\[ (x + 1)(7x - 6) = 0 \]

Step3: Find the zeros

To find the zeros, we set each factor equal to zero and solve for \(x\):
- For \((x + 1)=0\), we get \(x = -1\).
- For \((7x - 6)=0\), we solve for \(x\):
\[ 7x - 6 = 0 \]
\[ 7x = 6 \]
\[ x = \frac{6}{7} \]

Step4: Identify the correct choices

From the factoring, we have the factors \((x + 1)\) and \((7x - 6)\). From the zeros, we have \(x = -1\) and \(x = \frac{6}{7}\). Wait, but the options include \(x = 1\)? Wait, no, let's check again. Wait, when we factored, we had \((x + 1)(7x - 6)=0\), so the zeros are \(x = -1\) and \(x=\frac{6}{7}\). But the options also have \(x = 1\)? Wait, maybe I made a mistake. Wait, let's re - check the factoring.

Wait, let's use the quadratic formula to verify. For \(ax^2+bx + c = 0\), \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 7\), \(b = 1\), \(c=-6\). Then \(b^2-4ac=(1)^2-4\times7\times(-6)=1 + 168 = 169\), \(\sqrt{169}=13\). Then \(x=\frac{-1\pm13}{14}\). So \(x=\frac{-1 + 13}{14}=\frac{12}{14}=\frac{6}{7}\) and \(x=\frac{-1-13}{14}=\frac{-14}{14}=-1\). So the zeros are \(x=-1\) and \(x = \frac{6}{7}\), and the factors are \((x + 1)\) and \((7x - 6)\). But the options include \(x = 1\)? Wait, maybe there was a mistake in the problem or in my factoring? Wait, no, let's check the original equation again. The original equation is \(7x^2+x - 4 = 2\), which is \(7x^2+x - 6 = 0\). Factoring: \(7x^2+7x-6x - 6=7x(x + 1)-6(x + 1)=(x + 1)(7x - 6)\). So the factors are \((x + 1)\) and \((7x - 6)\), zeros are \(x=-1\) and \(x=\frac{6}{7}\). But the options also have \(x = 1\)? Wait, maybe the question has a typo, but according to our factoring and solving, the correct factors are \((x + 1)\) and \((7x - 6)\), and the correct zeros are \(x=-1\) and \(x=\frac{6}{7}\). But the options given: let's list the options again:

Options:
- \((x + 1)\)
- \(x = 1\)
- \((7x - 6)\)
- \(x=\frac{6}{7}\)
- \(x = 4\)
- \(x=-\frac{6}{7}\)
- \((x - 4)\)
- \(x=-1\)
- \((7x + 6)\)

Wait, we have \((x + 1)\) (factor), \((7x - 6)\) (factor), \(x=-1\) (zero), \(x=\frac{6}{7}\) (zero). But the problem says "Select 4 choices". Wait, maybe I made a mistake in factoring? Let's try another way. Let's use the quadratic formula again. \(a = 7\), \(b = 1\), \(c=-6\). \(x=\frac{-1\pm\sqrt{1 + 168}}{14}=\frac{-1\pm13}{14}\). So \(x=\frac{12}{14}=\frac{6}{7}\) and \(x=\frac{-14}{14}=-1\). So the zeros are \(x=-1\) and \(x=\frac{6}{7}\), factors are \((x + 1)\) and \((7x - 6)\). But the options include \(x = 1\)? Wait, maybe the original equation was different? Wait, no, the user provided \(7x^2 + x - 4 = 2\). So according to this, the correct factors are \((x + 1)\) and \((7x - 6)\), and the correct zeros are \(x=-1\) and \(x=\frac{6}{7}\). But the problem says "Select 4 choices". Wait, maybe I missed something. Wait, let's check the options again. Oh! Wait, maybe I made a mistake in the sign when factoring. Wait, \(7x^2 + x - 6\). Let's try factoring as \((7x + a)(x + b)=7x^2+(7b + a)x+ab\). We need \(ab=-6\) and \(7b + a = 1\). Let's try \(a = -6\), \(b = 1\): \(7(1)+(-6)=1\). Yes! So \((7x - 6)(x + 1)=7x^2+7x-6x - 6=7x^2+x - 6\). So that's correct. So the factors are \((x + 1)\) and \((7x - 6)\), zeros are \(x=-1\) and \(x=\frac{6}{7}\). But the options include \(x=-1\), \(x=\frac{6}{7}\), \((x + 1)\), \((7x - 6)\). Wait, but the problem says "Select 4 choices". Wait, maybe the problem has a typo, but according to the calculation, the correct choices are \((x + 1)\), \((7x - 6)\), \(x=-1\), \(x=\frac{6}{7}\). But let's check the options again. The options are:

1. \((x + 1)\) - correct (factor)
2. \(x = 1\) - incorrect
3. \((7x - 6)\) - correct (factor)
4. \(x=\frac{6}{7}\) - correct (zero)
5. \(x = 4\) - incorrect
6. \(x=-\frac{6}{7}\) - incorrect
7. \((x - 4)\) - incorrect
8. \(x=-1\) - correct (zero)
9. \((7x + 6)\) - incorrect

So the four correct choices are \((x + 1)\), \((7x - 6)\), \(x=\frac{6}{7}\), \(x=-1\). Wait, but the problem says "Select 4 choices", so these four. But let's confirm again.

Answer:

The four correct choices are:

• \((x + 1)\)
• \((7x - 6)\)
• \(x=\frac{6}{7}\)
• \(x=-1\)