Factoring with Grouping

Factoring with Grouping exercise

• Establish the equation corresponding to the area of the façade of the building.

  Hints

  The area of a rectangle is given by its height multiplied by its width.

  Look at this rectangle representing the house wall:

  Solution

  This rectangle represents the house wall.

  We calculate the area by multiplying the height by the width of the rectangle:

  • The height is given by $2x+3$.
  • The width is given by $x+2$.

  So we multiply those binomials to get the corresponding area:

  $(2x+3)\times (x+2)$

  Given that the area of the rectangle is $45$ yards squared, we have that

  $(2x+3)\times(x+2)=45$

• Solve the equation describing the area of the building to be painted.

  Hints

  FOIL multiplication means:

  • Multiply the First.
  • Multiply the Outer.
  • Multiply the Inner.
  • Multiply the Last.

  If you multiply any term by zero the product is also zero.

  Solution

  First, we use the FOIL method to factor the left-hand side of the equation:

  • Multiply the First to get $2x^2$.
  • Multiply the Outer to get $3x$.
  • Multiply the Inner to get $4x$.
  • Multiply the Last to get $6$.

  Adding all those terms together, we have $2x^2+3x+4x+6=45$.

  Combining like terms gives us $2x^2+7x+6=45$.

  Then subtract $45$ from both sides of the equation to get $2x^2+7x-39=0$.

  Now we are looking for the factors of $2\times (-39)=-78$ which sum to $7$:

  • $-1\times 78=-78$ but $-1+78=-77$
  • ...
  • $6\times (-13)=-78$ but $6+(-13)=-7$
  • $-6\times 13=-78$ and $-6+13=7$ $~~~~$✓

  Now we rewrite $2x^2+7x-39=0$ as

  $(2x^2-6x)+(13x-39)=0$

  and factor both terms inside the parenthesis:

  $2x(x-3)+13(x-3)=0$

  Again factoring, we get $(2x+13)\times (x-3)=0$. The zero product property tells us that we get either $2x+13=0$ or $x-3=0$.

  $\begin{array}{rcr} 2x+13 & = & 0\\ \color{#669900}{-13} & &\color{#669900}{-13}\\ 2x & = & -13\\ \color{#669900}{\div2} & & \color{#669900}{\div2}\\ x & = & -6.5 \end{array}$

  This can't be a solution in our case because $x$ is a length and thus cannot be negative.

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

  This is the solution we want. So the part of the building to paint is 6 yds. high and 3 yds. wide.

• Find the mistakes in Vincent's calculations.

  Hints

  Here you see an example for the FOIL multiplication:

  $(x+4)\times (x+5)=...$

  • Multiply the First to get $x\times x=x^2$.
  • Multiply the Outer to get $x\times 5=5x$.
  • Multiply the Inner to get $4\times x=4x$.
  • Multiply the Last to get $4\times 5=20$.

  Adding all these terms together gives us

  $...x^2+9x+20$

  Keep in mind that you only can combine like terms; for example, $2x+3x=5x$, but you can't combine anymore terms in $2x+3x^2$.

  Solution

  The FOIL method for multiplying binomials:

  • First
  • Outer
  • Inner
  • Last

  Keeping the FOIL method in mind, let's have a look at some examples:

  Supermarket

  $\begin{array}{rclll} (3x+1)\times(2x+2)&=&3x\times 2x+3x\times 2+1\times 2x+1\times 2\\ &=&6x^2+6x+2x+2&|&\text{ combining like terms}\\ &=&6x^2+8x+2 \end{array}$

  School

  $\begin{array}{rclll} (4x+1)\times(x+5)&=&4x\times x+4x\times 5+1\times x+1\times 5\\ &=&4x^2+20x+x+5&|&\text{ combining like terms}\\ &=&4x^2+21x+5 \end{array}$

  City hall

  $\begin{array}{rclll} (x+2)\times(5x+1)&=&x\times 5x+x\times 1+2\times 5x+2\times 1\\ &=&5x^2+x+10x+2&|&\text{ combining like terms}\\ &=&5x^2+11x+2 \end{array}$

• Factor each trinomial.

  Hints

  You can check the factorization using the FOIL method.

  If you multiply the factors of the linear terms of each binomial you get the factor of the resulting quadratic term of the trinomial.

  Solution

  If you want to get a factor a trinomial,

  $ax^2+bx+c=0$,

  you have to find the factors of $a\times c$ which sum to $b$.

  The other way is to check if a given factorization is correct using the FOIL method.

  Let's start with $3x^2+4x-4=0$. We need to find all factors of $3\times (-4)=-12$ and choose the pair which sums to $4$.

  • $-1\times 12=-12$ but $-1+12=-11$
  • ...
  • $2\times(-6)=-12$ but $2+(-6)=-4$
  • $-2\times 6=-12$ and $-2+6=4$.

  This gives us

  $3x^2+4x-4=(3x^2+6x)-(2x+4)=3x(x+2)-2(x+2)=(3x-2)\times(x+2)$.

  Let's try just another example: $-6x^2+2x+4=0$. Find all factors of $-6\times 4=-24$ and choose the pair which sum to $2$. We have that $-4\times 6=-24$ and $-4+6=2$, so we have:

  $-6x^2+2x+4=-6x^2-4x+6x+4=-2x(3x+2)+2(3x+2)=(-2x+2)\times(3x+2)$.

  The other way is to use the FOIL method:

  $(4x+4)\times(2x-3)=4x\times 2x+4x\times (-3)+4\times 2x+4\times (-3)$.

  Combining like terms we get $...=8x^2-12x+8x-12=8x^2-4x-12$.

  Last but not least we look at $(-x+3)\times(3x-1)=-x\times 3x-x\times (-1)+3\times 3x+3\times (-1)$. We can simplify this by combining like terms:

  $...=-3x^2+x+9x-3=-3x^2+10x-3$.

• Determine the solutions and decide which solution is a reasonable length.

  Hints

  Each factor gives a solution.

  Check each solution, keeping in mind that $x$ represents a length.

  Solution

  We start with the equation

  $(2x+13)\times (x-3)=0$.

  By the zero product property, we get that either $2x+13=0$ or $x-3=0$.

  $\begin{array}{rcr} 2x+13 & = & ~0\\ \color{#669900}{-13} & &\color{#669900}{-13}\\ 2x & = & ~-13\\ \color{#669900}{\div2} & & \color{#669900}{\div2}\\ x & = & ~-6.5 \end{array}$

  This can't be a solution, because $x$ is a length and thus cannot be negative.

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

  This is the desired solution. The part of the house wall to paint is 6 yds. high and 3 yds. wide.

• Determine the solutions of the equation.

  Hints

  Use FOIL multiplication to factor:

  • Multiply the First.
  • Multiply the Outer.
  • Multiply the Inner.
  • Multiply the Last.

  You have to find the factors of $-114$ which sum to $13$.

  Check your solutions (one is a decimal number) by inserting them into the equation above.

  Solution

  First, we use FOIL multiplication to factor:

  • Multiply the First to get $2x\times x=2x^2$.
  • Multiply the Outer to get $2x\times 4=8x$.
  • Multiply the Inner to $5x$.
  • Multiply the Last to get $5\times 4=20$.

  Adding all of these terms together, we get $2x^2+13x+20=77$.

  Subtracting $77$ on both sides gives us $2x^2+13x-57=0$.

  Now we are looking for the factors of $2\times (-57)=-114$ which sum to $13$:

  $-6\times 19=-114$ and $-6+19=13$. $~~~~$✓

  Now we can proceed by rewriting the equation above as

  $(2x^2-6x)+(19x-57)=2x(x-3)+19(x-3)=(2x+19)\times (x-3)=0$.

  Because a product equals zero if one of the factors equals zero we get either $2x+19=0$ or $x-3=0$.

  $~$
  $\begin{array}{rcr} 2x+19 & = & ~0\\ \color{#669900}{-19} & &\color{#669900}{-19}\\ 2x & = & ~-19\\ \color{#669900}{\div2} & & \color{#669900}{\div2}\\ x & = & ~-9.5 \end{array}$

  or

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

  $~$
  Those are the wanted solutions: Either $x=-9.5$ or $x=3$.