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To solve quadratic equations by factoring, we must make use of the zero-factor property.
Solve the following quadratic equations. (We will show the check for problem 1.)
$$\begin{array}{lllllllll}{x}^{2}-7x+12\hfill & =\hfill & 0.\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \begin{array}{l}\text{The\hspace{0.17em}equation\hspace{0.17em}is\hspace{0.17em}already\hspace{0.17em}}\\ \text{set\hspace{0.17em}equal\hspace{0.17em}to\hspace{0.17em}0}\text{.\hspace{0.17em}Factor}\text{.}\end{array}\hfill \\ \left(x-3\right)\left(x-4\right)\hfill & =\hfill & 0\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \text{Set\hspace{0.17em}each\hspace{0.17em}factor\hspace{0.17em}equal\hspace{0.17em}to\hspace{0.17em}0}\text{.}\hfill \\ \hfill x-3& =\hfill & 0\hfill & \hfill & \text{or}\hfill & \hfill & x-4\hfill & =\hfill & 0\hfill \\ \hfill x& =\hfill & 3\hfill & \hfill & \text{or}\hfill & \hfill & \hfill x& =\hfill & 4\hfill \end{array}$$
$\begin{array}{llllll}Check:\text{\hspace{0.17em}}\text{If}\text{\hspace{0.17em}}x=3,\text{\hspace{0.17em}}{x}^{2}-7x\hfill & +\hfill & 12\hfill & =\hfill & 0\hfill & \hfill \\ \hfill {3}^{2}-7\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}3& +\hfill & 12\hfill & =\hfill & 0\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill 9-21& +\hfill & 12\hfill & =\hfill & 0\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & 0\hfill & =\hfill & 0\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill \end{array}$
$\begin{array}{llllll}Check:\text{\hspace{0.17em}}\text{If}\text{\hspace{0.17em}}x=4,\text{\hspace{0.17em}}{x}^{2}-7x\hfill & +\hfill & 12\hfill & =\hfill & 0\hfill & \hfill \\ \hfill {4}^{2}-7\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}4& +\hfill & 12\hfill & =\hfill & 0\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill 16-28& +\hfill & 12\hfill & =\hfill & 0\hfill & \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\hfill \\ \hfill & \hfill & 0\hfill & =\hfill & 0\hfill & \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\hfill \end{array}$
Thus, the solutions to this equation are
$x=3,\text{\hspace{0.17em}}4.$
$$\begin{array}{lllll}\hfill {x}^{2}& =\hfill & 25.\hfill & \hfill & \text{Set\hspace{0.17em}the\hspace{0.17em}equation\hspace{0.17em}equal\hspace{0.17em}to\hspace{0.17em}}0.\hfill \\ \hfill {x}^{2}-25& =\hfill & 0\hfill & \hfill & \text{Factor}\text{.}\hfill \\ \left(x+5\right)\left(x-5\right)\hfill & =\hfill & 0\hfill & \hfill & \text{Set\hspace{0.17em}each\hspace{0.17em}factor\hspace{0.17em}equal\hspace{0.17em}to\hspace{0.17em}}0.\hfill \\ x+5=0\hfill & \text{or}\hfill & \hfill & x-5=0\hfill & \hfill \\ x=-5\hfill & \text{or}\hfill & \hfill & x=5\hfill & \hfill \end{array}$$
Thus, the solutions to this equation are
$x=5,-5.$
$$\begin{array}{lllll}\hfill {x}^{2}& =\hfill & 2x.\hfill & \hfill & \text{Set\hspace{0.17em}the\hspace{0.17em}equation\hspace{0.17em}equal\hspace{0.17em}to\hspace{0.17em}}0.\hfill \\ {x}^{2}-2x\hfill & =\hfill & 0\hfill & \hfill & \text{Factor}\text{.}\hfill \\ x\left(x-2\right)\hfill & \hfill & \hfill & \hfill & \text{Set\hspace{0.17em}each\hspace{0.17em}factor\hspace{0.17em}equal\hspace{0.17em}to\hspace{0.17em}}0.\hfill \\ x=0\hfill & \text{or}\hfill & \hfill & x-2=0\hfill & \hfill \\ \hfill & \hfill & \hfill & x=2\hfill & \hfill \end{array}$$
Thus, the solutions to this equation are
$x=0,\text{\hspace{0.17em}}2.$
$$\begin{array}{lllll}2{x}^{2}+7x-15\hfill & =\hfill & 0.\hfill & \hfill & \text{Factor}\text{.}\hfill \\ \left(2x-3\right)\left(x+5\right)\hfill & =\hfill & 0\hfill & \hfill & \text{Set\hspace{0.17em}each\hspace{0.17em}factor\hspace{0.17em}equal\hspace{0.17em}to\hspace{0.17em}}0.\hfill \\ 2x-3=0\hfill & \text{or}\hfill & \hfill & x+5=0\hfill & \hfill \\ 2x=3\hfill & \text{or}\hfill & \hfill & x=-5\hfill & \hfill \\ x=\frac{3}{2}\hfill & \hfill & \hfill & \hfill & \hfill \end{array}$$
Thus, the solutions to this equation are
$x=\frac{3}{2},-5.$
$63{x}^{2}=13x+6$
$$\begin{array}{lllll}63{x}^{2}-13x-6\hfill & =\hfill & 0\hfill & \hfill & \hfill \\ \left(9x+2\right)\left(7x-3\right)\hfill & =\hfill & 0\hfill & \hfill & \hfill \\ 9x+2=0\hfill & \hfill & \text{or}\hfill & \hfill & 7x-3=0\hfill \\ 9x=-2\hfill & \hfill & \text{or}\hfill & \hfill & 7x=3\hfill \\ x=\frac{-2}{9}\hfill & \hfill & \text{or}\hfill & \hfill & x=\frac{3}{7}\hfill \end{array}$$
Thus, the solutions to this equation are
$x=\frac{-2}{9},\frac{3}{7}.$
Solve the following equations, if possible.
$\left(x-7\right)\left(x+4\right)=0$
$x=7,\text{\hspace{0.17em}}-4$
$\left(2x+5\right)\left(5x-7\right)=0$
$x=\frac{-5}{2},\frac{7}{5}$
Let’s consider problems 4 and 5 of Sample Set A in more detail. Let’s look particularly at the factorizations $\left(2x-3\right)\left(x+5\right)=0$ and $\left(9x+2\right)\left(7x-3\right)=0.$ The next step is to set each factor equal to zero and solve. We can solve mentally if we understand how to solve linear equations: we transpose the constant from the variable term and then divide by the coefficient of the variable.
Solve the following equation mentally.
$\left(2x-3\right)\left(x+5\right)=0$
$$\begin{array}{lllll}2x-3\hfill & =\hfill & 0\hfill & \hfill & \text{Mentally\hspace{0.17em}add\hspace{0.17em}3\hspace{0.17em}to\hspace{0.17em}both\hspace{0.17em}sides}\text{.\hspace{0.17em}The\hspace{0.17em}constant\hspace{0.17em}changes\hspace{0.17em}sign}\text{.}\hfill \\ \hfill 2x& =\hfill & 3\hfill & \hfill & \begin{array}{l}\text{Divide\hspace{0.17em}by\hspace{0.17em}2,\hspace{0.17em}the\hspace{0.17em}coefficient\hspace{0.17em}of\hspace{0.17em}}x\text{.\hspace{0.17em}The\hspace{0.17em}2\hspace{0.17em}divides\hspace{0.17em}the\hspace{0.17em}constant\hspace{0.17em}3\hspace{0.17em}into\hspace{0.17em}}\frac{3}{2}\text{.\hspace{0.17em}}\\ \text{The\hspace{0.17em}coefficient\hspace{0.17em}becomes\hspace{0.17em}the\hspace{0.17em}denominator}\text{.}\end{array}\hfill \\ \hfill x& =\hfill & \frac{3}{2}\hfill & \hfill & \hfill \\ \hfill x+5& =\hfill & 0\hfill & \hfill & \text{Mentally\hspace{0.17em}subtract\hspace{0.17em}5\hspace{0.17em}from\hspace{0.17em}both\hspace{0.17em}sides}\text{.\hspace{0.17em}The\hspace{0.17em}constant\hspace{0.17em}changes\hspace{0.17em}sign}\text{.}\hfill \\ \hfill x& =\hfill & -5\hfill & \hfill & \text{Divide\hspace{0.17em}by\hspace{0.17em}the\hspace{0.17em}coefficient\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}}x\text{,\hspace{0.17em}1}\text{.The\hspace{0.17em}coefficient\hspace{0.17em}becomes\hspace{0.17em}the\hspace{0.17em}denominator}\text{.}\hfill \\ \hfill x=\frac{-5}{1}& =\hfill & -5\hfill & \hfill & \hfill \\ \hfill x& =\hfill & -5\hfill & \hfill & \hfill \end{array}$$
Now, we can immediately write the solution to the equation after factoring by looking at each factor, changing the sign of the constant, then dividing by the coefficient.
Solve $\left(9x+2\right)\left(7x-3\right)=0$ using this mental method.
$x=-\frac{2}{9},\frac{3}{7}$
For the following problems, solve the equations, if possible.
$\left(x+1\right)\left(x+3\right)=0$
$x=-1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3$
$\left(x+4\right)\left(x+9\right)=0$
$\left(x-5\right)\left(x-1\right)=0$
$x=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}5$
$\left(x-6\right)\left(x-3\right)=0$
$\left(x-4\right)\left(x+2\right)=0$
$x=-2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}4$
$\left(x+6\right)\left(x-1\right)=0$
$\left(2x+1\right)\left(x-7\right)=0$
$x=-\frac{1}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}7$
$\left(3x+2\right)\left(x-1\right)=0$
$\left(4x+3\right)\left(3x-2\right)=0$
$x=-\frac{3}{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{2}{3}$
$\left(5x-1\right)\left(4x+7\right)=0$
$\left(6x+5\right)\left(9x-4\right)=0$
$x=-\frac{5}{6},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{4}{9}$
$\left(3a+1\right)\left(3a-1\right)=0$
$x\left(x+4\right)=0$
$x=-4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0$
$y\left(y-5\right)=0$
$y\left(3y-4\right)=0$
$y=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{4}{3}$
$b\left(4b+5\right)=0$
$x\left(2x+1\right)\left(2x+8\right)=0$
$x=-4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\frac{1}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}0$
$y\left(5y+2\right)\left(2y-1\right)=0$
${\left(x-2\right)}^{2}=0$
${\left(a+1\right)}^{2}=0$
$x{\left(x-4\right)}^{2}=0$
$x=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}4$
$y{\left(y+9\right)}^{2}=0$
$y{\left(y-7\right)}^{2}=0$
$y=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}7$
$y{\left(y+5\right)}^{2}=0$
${x}^{2}-4=0$
$x=-2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}2$
${x}^{2}+9=0$
${x}^{2}-25=0$
${a}^{2}-100=0$
$a=-10,\text{\hspace{0.17em}}\text{\hspace{0.17em}}10$
${a}^{2}-81=0$
${b}^{2}-49=0$
$b=7,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-7$
${y}^{2}-1=0$
$3{a}^{2}-75=0$
$a=5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-5$
$5{b}^{2}-20=0$
${y}^{3}-y=0$
$y=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-1$
${a}^{2}=9$
${b}^{2}=4$
$b=2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2$
${b}^{2}=1$
${a}^{2}=36$
$a=6,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-6$
$3{a}^{2}=12$
$-2{x}^{2}=-4$
$x=\sqrt{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\sqrt{2}$
$-2{a}^{2}=-50$
$-7{b}^{2}=-63$
$b=3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3$
$-2{x}^{2}=-32$
$3{b}^{2}=48$
$b=4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-4$
${a}^{2}-8a+16=0$
${y}^{2}+9y+16=0$
${a}^{2}+6a+9=0$
${x}^{2}+12x=-36$
$3{a}^{2}+18a+27=0$
$2{m}^{3}+4{m}^{2}+2m=0$
$m=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-1$
$3m{n}^{2}-36mn+36m=0$
${a}^{2}+2a-3=0$
$a=-3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}1$
${a}^{2}+3a-10=0$
${x}^{2}+9x+14=0$
$x=-7,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2$
${x}^{2}-7x+12=3$
${b}^{2}+12b+27=0$
$b=-9,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3$
${b}^{2}-3b+2=0$
${x}^{2}-13x=-42$
$x=6,\text{\hspace{0.17em}}\text{\hspace{0.17em}}7$
${a}^{3}=-8{a}^{2}-15a$
$6{a}^{2}+13a+5=0$
$a=-\frac{5}{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\frac{1}{2}$
$6{x}^{2}-4x-2=0$
$12{a}^{2}+15a+3=0$
$a=-\frac{1}{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}-1$
$18{b}^{2}+24b+6=0$
$4{x}^{2}-4x=-1$
$2{x}^{2}=x+15$
$x=-\frac{5}{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}3$
$4{a}^{2}=4a+3$
$9{y}^{2}=9y+18$
( [link] ) Simplify ${\left({x}^{4}{y}^{3}\right)}^{2}{\left(x{y}^{2}\right)}^{4}.$
${x}^{12}{y}^{14}$
( [link] ) Write ${\left({x}^{-2}{y}^{3}{w}^{4}\right)}^{-2}$ so that only positive exponents appear.
( [link] ) Find the sum: $\frac{x}{{x}^{2}-x-2}+\frac{1}{{x}^{2}-3x+2}.$
$\frac{{x}^{2}+1}{\left(x+1\right)\left(x-1\right)\left(x-2\right)}$
( [link] ) Simplify $\frac{\frac{1}{a}+\frac{1}{b}}{\frac{1}{a}-\frac{1}{b}}.$
( [link] ) Solve $\left(x+4\right)\left(3x+1\right)=0.$
$x=-4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{-1}{3}$
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