Factoring by Grouping

Factoring by Grouping exercise

• Explain how to factor the given quadratic polynomial by grouping.

  Hints

  To factor quadratic polynomials by grouping, first multiply $a$ with $c$. Then sum all pairs of factors of this product and find the pair which sums to $b$.

  Working backwards, we have:

  $\begin{array}{rcl} (3x-1)(x+2)&=&3x(x+2)-(x+2)\\ &=&3x^2+6x-x-2 \end{array}$

  Solution

  In order to factor such a quadratic polynomial,

  $ax^2+bx+c$,

  with $a$, $b$, and $c$ as coefficients, we must find all pairs of factors of $a\times c$.

  Then we search for the pair of factors whose sum is $b$.

  Let's have a look at an example: $3x^2+5x-2$.

  Here we have $a=3$, $b=5$, and $c=-2$ and thus $a\times c=3\times -2=-6$.

  The pairs of factors of $-6$ and their corresponding sums are:

  $\begin{array}{rr|r} \text{factor}&\text{factor}&\text{sum}\\ \hline -1&6&5\\ 1&-6&-5\\ -2&3&1\\ 2&-3&-1 \end{array}$

  The pair of factors which sums to $5$ is $-1$ and $6$.

  The following grouping

  $3x^2+5x-2=3x^2+6x-x-2$

  then gives us

  $3x(x+2)-(x+2)=(3x-1)(x+2)$.

  which is the factorization we are looking for.

• Factor the polynomial $15x^2+9x-6$.

  Hints

  The product of $15$ and $-6$ is $-90$.

  You can write $-90$ as a product of two factors. Find the pair of factors that add up to $9$.

  In the example,

  $\begin{array}{rcl} (2x+3)(4x-2)&=&2x(4x-2)+3(4x-2)\\ &=&8x^2-4x+12x-6\\ &=&8x^2+8x-6 \end{array}$

  we have that $8\times -6=48$, as well as $-4\times 12=48$. We can see that $-4+12=8$.

  First multiply $a$ and $b$. Then we check all pairs of factors of $ab$ whose product is $ab$. If we add a pair of factors together and get $b$, then we've found the pair we are looking for.

  Solution

  To factor the quadratic trinomial $15x^2+9x-6$ with $a=15$, $b=9$, and $c=-6$, we first look for the pairs of factors of $a\times c$ that sum to $b$.

  We have that $a\times c=-90$, so we investigate all pairs of factors of $-90$:

  ${\small\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c} \text{Factors of }-90&-1,90&1,-90&-2,45&2,-45&-3,30&3,-30&-5,18&5,-18&-6,15&6,-15 \\ \hline \text{Sum of factors }&89&-89&43&-43&27&-27&13&-13&9&-9 \end{array}}$

  The sum of the factors $-6$ and $15$ is $-6+15=9$. So this is the pair we want.

  Next we group the quadratic polynomial as follows:

  $15x^2+9x-6=15x^2-6x+15x-6$.

  Grouping the binomials together, we get $(15x^2-6x)+(15x-6)$.

  The greatest common factor of the left binomial is $3x$ and of the right binomial is $3$:

  $(15x^2-6x)+(15x-6)=3x(5x-2)+3(5x-2)$.

  Now we see that both summands have the binomial $5x-2$ as a common factor, and thus we can factor even further:

  $3x(5x-2)+3(5x-2)=(3x+3)(5x-2)$.

• Indicate the coefficients of the quadratic polynomials.

  Hints

  Keep in mind that the $-$ sign belongs to the coefficient.

  For the polynomial $3x^2+2x+4$, we have that $a=3$, $b=2$, and $c=4$.

  Pay attention: Neiter $x^2$ nor $x$ belong to $a$ or $b$.

  Solution

  The standard form of quadratic polynomial is given by

  $ax^2+bx+c$, where $a\neq 1$.

  Before we start to group quadratic polynomials we have to note the values of $a$, $b$, and $c$.

  For the given polynomials, we look for the factor of the quadratic term $x^2$, $a$, the linear term, $b$, and the constant term, $c$:

  • For $6x^2+x-2$, we have $a=6$, $b=1$, and $c=-2$.
  • For $9x^2-25x-6$, we have $a=9$, $b=-25$, and $c=-6$.
  • For $8x^2+4x-12$, we have $a=8$, $b=4$, and $c=-12$.
  • For $8x^2+14x+3$, we have $a=8$, $b=14$, and $c=3$.

• Factor the given polynomial.

  Hints

  $ax^2+bx+c$ is the standard form of a quadratic polynomial.

  For example, with $x^2-x+1$, we have that $a=1$, $b=-1$, and $c=1$.

  Look for the pair of factors of $a\times c$ that sum to $b$.

  Once you group a polynomial into two binomials, each binomial has a greatest common factor. So you can factor the greatest common factor from each binomial.

  Solution

  To factorize the trinomial $6x^2+x-2$ we first look at the coefficients: Here we have $a=6$, $b=1$ and $c=-2$.

  We're looking for the pairs of factors of $6\times (-2)=-12$ which sum to $1$.

  $\begin{array}{rr|r} \text{factor}&\text{factor}&\text{sum}\\ \hline -1&12&11\\ 1&-12&-11\\ -2&6&4\\ 2&-6&-4\\ -3&4&1 \end{array}$

  Now we can stop: We've found the factors $-3$ and $4$.

  We can rewrite $6x^2+x-2$ as follows:

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

  Just look at the two resulting binomials in parentheses. The greatest common factor of the first one is $3x$ and of the second one is $2$:

  $(6x^2-3x)+(4x-2)=3x(2x-1)+2(2x-1)$.

  Now we see that both summands have the binomial $2x-1$ as a factor in common. So we factor to get, $3x(2x-1)+2(2x-1)=(3x+2)(2x-1)$.

  The side lengths Adventure Mike was looking for are then $3x+2$ and $2x-1$.

• Complete the following table of factors and sums.

  Hints

  The product of a pair of factors must be $-90$. If you know one factor, just divide $-90$ by that factor to get the other factor in the pair.

  An example of a pair of factors of $-90$ and their sum:

  Solution

  For $15x^2+9x-6$, we first multiply $15\times -6=-90$.

  We then know that we are looking for all pairs of factors of $-90$, and one pair in particular which sum up to $9$.

  Let's calculate the products and sums of all pairs of factors of $-90$:

  • $-1\times 90=-90$ and $-1+90=89$
  • $1\times (-90)=-90$ and $1-90=-89$
  • $-2\times 45=-90$ and $-1+90=89$
  • $2\times (-45)=-90$ and $2-45=-43$
  • $-3\times 30=-90$ and $-3+30=27$
  • $3\times (-30)=-90$ and $3-30=-27$
  • $-5\times 18=-90$ and $-5+18=13$
  • $5\times (-18)=-90$ and $5-18=-13$
  • $-6\times 15=-90$ and $-6+15=9$
  • $6\times (-15)=-90$ and $6-15=-9$

  We can now see that the pair $-6$ and $15$ satisfies our requirements: $-6\times 15=-90$ as well as $-6+15=9$.

• Consider the appropriate factorization.

  Hints

  You can either factor by grouping or use the FOIL method for binomials.

  An example of the FOIL method:

  $\begin{array}{rcl} (5x+1)(x+5)&=&5x^2+5x(5)+1x+1(5)\\ &=&5x^2+26x+5 \end{array}$

  Remember: First, Outer, Inner, Last.

  If you'd like to factor $ax^2+bx+c$ by grouping then:

  • Find all pairs of factors of $a\times c$.
  • Check which pair sums to $b$.

  Solution

  To match the quadratic polynomial in standard form with its factorization, you can either factor the quadratic polynomial or multiply the binomials.

  Let's go over how to use both methods with the polynomial $2x^2-11x+12$.

  Factoring the Polynomial:

  First we note that $a=2$, $b=-11$ and $c=12$.

  First multiply $a$ and $c$ to get $2\times 12=24$.

  Next look for the pair of factors of $24$ which sum to $-11$; the pair we are looking for is $-8$ and $-3$.

  Then group to get $2x^2-11x+12=(2x^2-8x)-(3x-12)=2x(x-4)-3(x-4)$.

  Then factor out the greatest common factors of the binomials to get $2x(x-4)-3(x-4)=(2x-3)(x-4)=(-2x+3)(-x+4)$.

  Lastly, multiply each binomial with $-1$.

  Multiplying the Binomials:

  You can also use FOIL multiplication with $(3x-4)(2x-2)$.

  We have that $(3x-4)(2x-2)=6x^2-6x-8x+8=6x^2-14x+8$.