Factorising trinomials: ax2+bx+cax^2 + bx + c

Quadratic equations are also polynomial expressions which are called such because the variable is squared. It has the general formula ax2+bx+c=0ax^2 + bx + c = 0, but more often than not, it comes in many forms where the terms are rearranged and grouped. You can either just stick to what form you see, or rearrange it so it looks like the general formula, then you can start factoring them.

These equations are very easy to factor, given of course that you remember our lessons about factoring polynomial expressions from previous chapter, and if you would utilize a free factoring calculator online. We learned that Polynomials can be factored by looking at the common factors, which is why in this chapter, we will also do the same.

We also learned that factoring can also be done by grouping, like in the case of x37x2+2x14x^3-7x^2+2x-14 where we group the first two terms and the last two terms to get the factors, x - 7 and x2+2x^2 + 2. You can either group the first two terms and the last two terms, or group them like how you want to and you would still arrive at the same answer.

In this chapter, apart from reviewing the concepts we learned from previous chapter, we will also get to learn about the factoring the differences of squares and factoring trinomials. Differences of squares can be mathematically illustrated as (a)2(b)2(a)^2 - (b)^2 while trinomials are defined mathematically as a polynomial expression that has three terms.

In factoring the differences of squares, using the given general form (a)2(b)2(a)^2 - (b)^2, we know that it has the factors (a+b) and (a-b), so in factoring the differences of squares, we need to express the given equation into the form (a)2(b)2(a)^2 - (b)^2 and factor them by filling in the expression inside the two parenthesis ( _ + _) and ( _ -_ ).

In factoring trinomials, we use the “cross-multiply then check” method. So if we’re given the trinomial b2b20b^2-b-20, we know that the factors of b2b^2 are b and b, and the factors of 20 can be 10 and 2, 5 and 4, and 20 and 1, so we start there. Now in order for us to know which factor of 20 would be used to complete the expression ( b + _ ) (b - _) we cross multiply, the factors of the first term and the last term then check if they’re equal to the middle term, -b.

Factorising trinomials: ax2+bx+cax^2 + bx + c


  • 2.
    • c)
      2x314x2+24x2{x^3} - 14{x^2} + 24x
    • d)
      14+5yy214 + 5y - {y^2}
  • 3.
    • b)
      5x2+8x+35{x^2} + 8x + 3
    • d)
      6m213m86{m^2} - 13m - 8
    • f)
      63+20z3z263 + 20z - 3{z^2}
    • g)
      8x2+8x68{x^2} + 8x - 6
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Factorising trinomials: ax2+bx+cax^2 + bx + c

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