Question

use the zero product property to find the solutions to the equation $6x^2 - 5x = 56$
$\bigcirc$ $x = -8$ or $x = \frac{7}{8}$
$\bigcirc$ $x = -\frac{8}{3}$ or $x = \frac{7}{2}$
$\bigcirc$ $x = -\frac{1}{2}$ or $x = \frac{56}{3}$
$\bigcirc$ $x = -\frac{3}{7}$ or $x = \frac{2}{3}$

Explanation:

Step1: Rewrite the equation

First, we need to rewrite the quadratic equation \(6x^{2}-5x = 56\) in standard form \(ax^{2}+bx + c=0\). Subtract 56 from both sides:
\(6x^{2}-5x - 56=0\)

Step2: Factor the quadratic

We factor the quadratic \(6x^{2}-5x - 56\). We need two numbers that multiply to \(6\times(- 56)=-336\) and add up to \(-5\). The numbers are \(-21\) and \(16\) since \(-21\times16=-336\) and \(-21 + 16=-5\).
Rewrite the middle term:
\(6x^{2}+16x-21x - 56 = 0\)
Group the terms:
\((6x^{2}+16x)-(21x + 56)=0\)
Factor out the greatest common factor from each group:
\(2x(3x + 8)-7(3x + 8)=0\)
Now, factor out \((3x + 8)\):
\((2x - 7)(3x+8)=0\) Wait, no, wait. Wait, let's check again. Wait, \(6x^{2}+16x-21x - 56\), factoring \(6x^{2}+16x = 2x(3x + 8)\) and \(-21x-56=-7(3x + 8)\). So the factored form is \((2x - 7)(3x + 8)=0\)? Wait, no, \(2x\times3x = 6x^{2}\), \(2x\times8 = 16x\), \(-7\times3x=-21x\), \(-7\times8=-56\). Yes. Wait, but let's use zero - product property. The zero - product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
Wait, maybe I made a mistake in factoring. Let's try another way. Let's use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here, \(a = 6\), \(b=-5\), \(c=-56\).
\(x=\frac{5\pm\sqrt{(-5)^{2}-4\times6\times(-56)}}{2\times6}=\frac{5\pm\sqrt{25 + 1344}}{12}=\frac{5\pm\sqrt{1369}}{12}=\frac{5\pm37}{12}\)
Case 1: \(x=\frac{5 + 37}{12}=\frac{42}{12}=\frac{7}{2}\)? Wait, no, that's not matching. Wait, maybe my factoring was wrong. Wait, let's go back to factoring.
We have \(6x^{2}-5x - 56\). Let's try to factor by grouping correctly. We need two numbers \(m\) and \(n\) such that \(m\times n=6\times(-56)=-336\) and \(m + n=-5\). Let's list the factor pairs of \(-336\):
\(-336=21\times(-16)\), \(21+(-16) = 5\), not \(-5\). \(-336=-21\times16\), \(-21 + 16=-5\). Yes. So we rewrite the middle term as \(-21x+16x\) (wait, no, the original middle term is \(-5x\), so we need \(m + n=-5\). So \(m=-21\), \(n = 16\). So \(6x^{2}-21x+16x - 56\). Now group:
\((6x^{2}-21x)+(16x - 56)=0\)
Factor out \(3x\) from the first group and \(8\) from the second group:
\(3x(2x - 7)+8(2x - 7)=0\)
Ah! There we go. So the factored form is \((3x + 8)(2x - 7)=0\) (wait, \(3x\times2x=6x^{2}\), \(3x\times(-7)=-21x\), \(8\times2x = 16x\), \(8\times(-7)=-56\)). Yes, so \((3x + 8)(2x - 7)=0\) is wrong, it's \((3x + 8)(2x - 7)=0\)? Wait, no, \(3x\times2x = 6x^{2}\), \(3x\times(-7)=-21x\), \(8\times2x=16x\), \(8\times(-7) = - 56\). So combining like terms \(-21x + 16x=-5x\). So the factored form is \((3x + 8)(2x - 7)=0\)? Wait, no, \((3x + 8)(2x - 7)=6x^{2}-21x + 16x-56=6x^{2}-5x - 56\). Yes!
Now, apply the zero - product property:
If \((3x + 8)(2x - 7)=0\), then either \(3x+8 = 0\) or \(2x - 7=0\)
For \(3x+8 = 0\), subtract 8 from both sides: \(3x=-8\), then \(x =-\frac{8}{3}\)
For \(2x - 7=0\), add 7 to both sides: \(2x = 7\), then \(x=\frac{7}{2}\)

Wait, but let's check the options. The second option is \(x =-\frac{8}{3}\) or \(x=\frac{7}{2}\). Let's verify by plugging back into the original equation.
Original equation: \(6x^{2}-5x = 56\)
For \(x =-\frac{8}{3}\):
Left - hand side: \(6\times(-\frac{8}{3})^{2}-5\times(-\frac{8}{3})=6\times\frac{64}{9}+\frac{40}{3}=\frac{384}{9}+\frac{120}{9}=\frac{504}{9} = 56\), which matches the right - hand side.
For \(x=\frac{7}{2}\):
Left - hand side: \(6\times(\frac{7}{2})^{2}-5\times(\frac{7}{2})=6\times\frac{49}{4}-\frac{35}{2}=\frac{294}{4}-\frac{70}{4}=\frac{224}{4}=56\), which also matches.

Answer:

\(x =-\frac{8}{3}\) or \(x=\frac{7}{2}\) (the second option: \(x =-\frac{8}{3}\) or \(x=\frac{7}{2}\))