Solving Quadratic Equations by Factoring

Solving Quadratic Equations by Factoring exercise

• State the solutions of the equation $(x+6)(x-1)=0$.

  Hints

  Sure $4\times 0=0$ as well as $0\times 5=0$.

  Here you see an example for the solution of an equation.

  You can check your solutions by putting them into the starting equation.

  Solution

  The zero factor property says that if a product $a\times b$ is equal to zero, then one of its factors, either $a$ or $b$, is equal to zero.

  We can use this fact to solve factorized quadratic equations. For example,

  $(x+6)(x-1)=0$.

  By the zero factor property, we know that either $x+6=0$ or $x-1=0$. Using opposite operations we get

  $\begin{array}{rcl} x+6 & = & ~0\\ \color{#669900}{-6} & &\color{#669900}{-6}\\ x & = & ~-6 \end{array}$

  as well as

  $\begin{array}{rcl} x-1 & = & ~0\\ \color{#669900}{+1} & &\color{#669900}{+1}\\ x & = & ~1 \end{array}$

  So, we can conclude that the equation above has two solutions: $x=-6$ or $x=1$.

• Solve the quadratic equation.

  Hints

  To factor a quadratic term $ax^2+bx+c$, find the factors of $a\times c$ which sum to $b$.

  Use the zero factors property: if a product is zero, then one of its factors must be zero.

  To solve linear equations use opposite operations.

  Solution

  If we could factor the quadratic term $x^2+x-6=(x+m)(x+n)$, then we would be able to solve this equation using the zero factor property.

  With the reverse FOIL method we can determine $m$ and $n$ as follows:

  1. $(x+m)(x+n)=x^2+nx+mx+mn=x^2+(m+n)x+mn$.
  2. So we have to find the $m$ and $n$ which statisfy $m\times n=-6$ and $m+n=1$.
  3. Let's check:

  • $-1\times 6=-6$ but $-1+6=5$
  • $1\times (-6)=-6$ but $1-6=-5$
  • $-2\times 3=-6$ and $-2+3=1$ $~~~~~$✓

  Now we know that $x^2+x-6=(x-2)(x+3)$ and thus the equation above is equivalent to

  $(x-2)(x+3)=0$.

  Thus we have to solve $x+6=0$ and $x-1=0$. Using opposite operations we get

  $\begin{array}{rcl} x-2 & = & ~0\\ \color{#669900}{+2} & &\color{#669900}{+2}\\ x & = & ~2 \end{array}$

  as well as

  $\begin{array}{rcl} x+3 & = & ~0\\ \color{#669900}{-3} & &\color{#669900}{-3}\\ x & = & ~-3 \end{array}$

  Thus the solutions are $x=2$ or $x=-3$.

• Determine the solutions of the factorized equations.

  Hints

  Use the zero factors property: if a product is zero, then one of its factors must be zero.

  Each factor is linear. To solve a linear equation use opposite operations.

  You can solve an equation like $x^2-8x=0$ as follows:

  1. Factor out an $x$ to get $x(x-8)=0$.
  2. Thus either $x=0$ or $x-8=0$, which is equivalent to $x=8$.

  Solution

  $(x-1)(x-2)=0$: either $x-1=0$ or $x-2=0$. This leads to the following solutions

  $\begin{array}{rcl} x-1 & = & ~0\\ \color{#669900}{+1} & &\color{#669900}{+1}\\ x & = & ~1 \end{array}$

  or

  $\begin{array}{rcl} x-2 & = & ~0\\ \color{#669900}{+2} & &\color{#669900}{+2}\\ x & = & ~2 \end{array}$

  $~$

  $(2x+4)(x-5)=0$: either $2x+4=0$ or $x-5=0$. So we solve these equations to get

  $\begin{array}{rcl} 2x+4 & = & ~0\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ 2x & = & ~-4 \\ \color{#669900}{\div 2} & &\color{#669900}{\div 2}\\ x&=&~-2 \end{array}$

  or

  $\begin{array}{rcl} x-5 & = & ~0\\ \color{#669900}{+5} & &\color{#669900}{+5}\\ x & = & ~5 \end{array}$

  $~$

  $(x+7)(-x+3)=0$: either $x+7=0$ or $-x+3=0$. Again we use opposite operations to solve those equations

  $\begin{array}{rcl} x+7 & = & ~0\\ \color{#669900}{-7} & &\color{#669900}{-7}\\ x & = & ~-7 \end{array}$

  or

  $\begin{array}{rcl} -x+3 & = & ~0\\ \color{#669900}{+x} & &\color{#669900}{+x}\\ 3 & = & ~x \end{array}$

  $~$

  Let's have a look at a last example $x^2+4x=0$ which isn't factored yet. You can factor out $x$ to get $x(x+4)=0$. So either $x=0$ or $x+4=0$. We still have to solve the second equation:

  $\begin{array}{rcl} x+4 & = & ~0\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ x & = & ~-4 \end{array}$

• Multiply the two binomials using the FOIL method.

  Hints

  FOIL is a mnemonic device for multiplying binomials:

  • First
  • Outer
  • Inner
  • Last

  Here you see an example using FOIL:

  $(x-5)(x+2)=$

  • F: $x\times x=x^2$
  • O: $x\times 2=2x$
  • I: $-5\times x=-5x$
  • L: $-5\times 2=-10$

  Add and combine the like terms. For the example above we get

  $x^2+2x-5x-10=x^2-3x-10$.

  Solution

  FOIL is a mnemonic device for multiplying two binomials:

  First, Outer, Inner, Last.

  $~$

  $(2x-1)(x+1)=$

  • F: $2x\times x=2x^2$
  • O: $2x\times 1=2x$
  • I: $-1\times x=-x$
  • L: $-1\times 1=-1$

  Now we add the terms and combine the like terms to get $2x^2+2x-x-1=2x^2+x-1$.

  $(x-5)(2x+2)=$

  • F: $x\times 2x=2x^2$
  • O: $x\times 2=2x$
  • I: $-5\times 2x=-10x$
  • L: $-5\times 2=-10$

  We still have to add the terms to $2x^2+2x-10x-10$ and combine the like terms $2x^2-8x-10$.

  $(x+3)(x-3)=$

  • F: $x\times x=x^2$
  • O: $x\times (-3)=-3x$
  • I: $3\times x=3x$
  • L: $3\times (-3)=-9$

  Next add the terms $x^2-3x+3x-9$ and combine the like terms $x^2-9$.

  $(x-3)(x-3)=$

  • F: $x\times x=x^2$
  • O: $x\times (-3)=-3x$
  • I: $-3\times x=-3x$
  • L: $(-3)\times (-3)=9$

  Almost done: we just have to add the terms to get $x^2-3x-3x+9=x^2-6x+9$.

• Check the factorization.

  Hints

  FOIL is a mnemonic device for multiplying binomials:

  • First
  • Outer
  • Inner
  • Last

  Here you see an example for using FOIL.

  $2x$ and $x$ are like terms and can be combined to get $2x+x=3x$.

  Two terms are correct.

  Solution

  FOIL is a mnemonic device for multiplying binomials. The letters stand for

  • First
  • Outer
  • Inner
  • Last

  Let's have a look at the example beside.

  • F: $x\times x=x^2$
  • O: $x\times 3=3x$
  • I: $-2\times x=-2x$
  • L: $-2\times 3=-6$

  Adding these terms leads to

  $x^2+3x-2x-6$

  and combing the like terms gives

  $x^2+x-6$.

  So we have together

  $x^2+x-6=(x-2)(x+3)$.

• Determine the code.

  Hints

  To factor a quadratic term $x^2+bx+c$, find the factors of $c$ which sum to $b$.

  For the equation above, check all factors of $m\times n=-28$ and pick the pair $m$ and $n$ satisfying $m+n=-3$.

  If you know the factorization, for example $(x-2)(x+15)$, just solve the two equations:

  • $x-2=0$
  • $x+15=0$

  Check your solution: put both solutions for $x$ into the equation above to see if they satisfy it.

  Solution

  To solve this quadratic equation we factorize the quadratic term $x^2-3x-28$.

  For this we check the factors $m\times n=-28$ and choose those which sum to $m+n=-3$:

  • $m=-1$ and $n=28$ but $m+n=27$
  • $m=1$ and $n=-28$ but $m+n=-27$
  • $m=-2$ and $n=14$ but $m+n=12$
  • $m=2$ and $n=-14$ but $m+n=-12$
  • $m=-4$ and $n=7$ but $m+n=3$
  • $m=4$ and $n=-7$ and $m+n=-3$ $~~~~~$✓

  Thus $x^2-3x-28=(x+4)(x-7)$ and so the equation above is equivalent to

  $(x+4)(x-7)=0$.

  We conclude that either $x+4=0$ or $x-7=0$. Using opposite operations we get

  $\begin{array}{rcl} x+4 & = & ~0\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ x & = & ~-4 \end{array}$

  as well as

  $\begin{array}{rcl} x-7 & = & ~0\\ \color{#669900}{+7} & &\color{#669900}{+7}\\ x & = & ~7 \end{array}$

  Now we know Grandma Millers Code: $x=-4$ and $x=7$.