Content code
m1387
Slug (identifier)
product-sum-technique
Parent content
Grades
Secondary IV
Topic
Mathematics
Tags
trinôme
somme
product
nombres
mise en évidence
évidence double
factorisation
factoriser
factorisation d'un trinôme
somme produit
somme-produit
produit-somme
sp
méthode du produit somme
méthode somme produit
méthode produit somme
Content
Contenu
Content
Corps

The product-sum technique makes it possible to factor a trinomial of the form |ax^2+bx+c.|

Content
Corps

To factor a trinomial of the form |\color{green}{a}x^2 + \color{blue}{b}x + \color{green}{c}| with the product-sum technique, follow these steps.

  1. Find two numbers |\color{red}{m}| and |\color{red}{n}| whose product equals the value of |\color{green}{a}| multiplied by |\color{green}{c}|, and whose sum equals the value of |\color{blue}{b}|.
    ||\color{green}{\text{ Product }} =\color{green}{a}\color{green}{c}\ \ \ \ \ \ \color{blue}{\text{Sum }}=\color{blue}{b}||

  2. Decompose the term |bx| in the trinomial using the two numbers found.

  3. Factor by grouping.

Content
Corps

This method can be difficult if the values of |a|, |b|, and |c| are fractions or fairly large integers (both positive and negative).

Moreover, if the discriminant of the trinomial |ax^2+bx+c| is negative, it is not possible to factor the trinomial at all.||b^2-4ac<0 \ \Longrightarrow \text{not factorable}||

Content
Corps

Consider the trinomial |x^2 + 4x – 32.|

  1. Find the product and sum
    Identify the parameters |a|, |b|, and |c| of this trinomial.

    ||\color{green}{a} = \color{green}{1},\ \color{blue}{b} = \color{blue}{4},\ \color{green}{c} = \color{green}{-32}|| ||\begin{align}\color{green}{\text{Product}}&=\color{green}{a}\color{green}{c}& \color{blue}{\text{Sum}} &= \color{blue}{b}\\ &= \color{green}{1}\ \times\color{green}{-32}&&=\color{blue}{4} \\ &=\color{green}{-32}\end{align}||

    We are looking for two numbers with a product of |-32| and a sum of |4|. Use trial and error to determine them, like so:

    ||\begin{align}-1\times 32&=\color{green}{-32},\ \text{but}\ \ -1+32=31\\ 1\times -32&=\color{green}{-32}, \ \text{but}\ \ \ 1+(-32)=-31\\-2\times 16&=\color{green}{-32}, \ \text{but}\ \ -2+16=14\\ 2\times -16&=\color{green}{-32}, \ \text{but}\ \ \ 2+(-16)=-14\\ \color{red}{-4}\times \color{red}{8}&=\color{green}{-32}\ \ \ \text{and}\ \ \color{red}{-4}+\color{red}{8}=\color{blue}{4}\end{align}|| The two numbers are therefore |\color{red}{-4}| and |\color{red}{8}|.

  2. Decompose the term |bx| in the trinomial using the two numbers found

    ||\begin{align}x^2+4x-32&=x^2\color{red} {+4x}-32\\ &=x^2\color{red} {-4x+8x}-32\end{align}||

  3. Factor by grouping

    ||\begin{align}x^2+4x-32&=x^2-4x+8x-32\\&=x(x-4)+8(x-4)\\&=(x-4)(x+8)\end{align}||

Content
Corps

Consider the trinomial |6x^2+16x+8.|

  1. Find the product and sum
    Before trying the product-sum technique, notice the polynomial has common factors. It is possible to factor out a common factor. ||6x^2+16x+8\\ 2(3x^2+8x+4)|| Next, apply the product-sum technique to the trinomial |3x^2+8x+4|.

    Identify the parameters |a|, |b|, and |c| of the trinomial. ||\color{green}{a} = \color{green}{3}, \color{blue}{b} = \color{blue}{8}, \color{green}{c} = \color{green}{4}|| ||\begin{align}\color{green}{\text{Product}}&=\color{green}{a}\color{green}{c}& \color{blue}{\text{Sum}} &= \color{blue}{b}\\ &= \color{green}{3}\times \color{green}{4}&&=\color{blue}{8} \\ &=\color{green}{12}\end{align}|| Look for two numbers with a product of |12| and a sum of |8|.
    Use trial and error to determine them, as follows. ||\begin{align}1\times 12&=\color{green}{12},\ \text{but}\ \ 1+12=13\\ 3\times 4&=\color{green}{12}, \ \text{but}\ \ \ 3+4=7\\ \color{red}{2}\times \color{red}{6}&=\color{green}{12}\ \ \ \text{and}\ \ \color{red}{2}+\color{red}{6}=\color{blue}{8}\end{align}|| The two numbers are therefore |\color{red}{2}| and |\color{red}{6}|.

  2. Decompose the term |bx| in the trinomial using the two numbers found ||\begin{align}6x^2+16x+8&=2(3x^2\color{red} {+8x}+4)\\ &=2(3x^2\color{red} {+2x+6x}+4)\end{align}||

  3. Factor by grouping ||\begin{align}6x^2+16x+8&=2(3x^2+2x+6x+4)\\&=2\left(x(3x+2)+2(3x+2)\right)\\&=2(3x+2)(x+2)\end{align}||

Remove audio playback
No