Factoring trinomials worksheet

Factoring trinomials worksheet (solved examples) 

Factorise  the  following  Quadratic trinomials  \mathbf{ax^{2}+bx+c}
  by spiliting  the middle terms.  

We know, in order to factorise the expression ax2 + bx + c, we have to find two numbers p and q, such that p + q = b and p × q = ac.

Example 1: Factorise \mathbf{x^{2}-7x+12} .

Solution. Find two factors p and q such that p+q=-7 and pq =12 . Take p=-3 and q=-4  then 12= -3 x (-4) and -3-4=-7 .

Therefore  \mathbf{x^{2}-7x+12} = \mathbf{x^{2}+(-3-4)x+12}

=\mathbf{x^{2}-3x-4x +12}

=\mathbf{x(x-3)-4(x-3)}

=\mathbf{(x-3)(x-4)}

Example 2: Obtain the factors of \mathbf{x^{2}-4x-12} .

Solution.  \mathbf{x^{2}-4x-12} = \mathbf{x^{2}-6x+2x-12}

= \boldsymbol{x(x-6)+2(x-6)}

=\mathbf{(x+2)(x-6)}

Example 3: Find the factors of \mathbf{3x^{2}+9x+6}.

Solution. Note that 3 is common factor in all the terms. Therefore

\mathbf{3x^{2}+9x+6} = \mathbf{3(x^{2}+3x+2)} 

=\mathbf{3(x^{2}+2x+x+2)}

=\mathbf{3(x(x+2)+1(x+2))}

=\mathbf{3(x+1)(x+2)}

Example 4: Factorise \mathbf{x^{2}+5\sqrt{3}x+12} .

Solution.  We split 5\sqrt{3} into two parts whose sum is 5\sqrt{3} and product is 12. Clearly 4\sqrt{3}+\sqrt{3}= 5\sqrt{3} and 4\sqrt{3}\times \sqrt{3}= 12

\therefore x^{2}+5\sqrt{3}x+12= x^{2}+4\sqrt{3}x+\sqrt{3}x+12

=x(x+4\sqrt{3})+\sqrt{3}(x+4\sqrt{3})

=(x+4\sqrt{3})(x+\sqrt{3})

Hence \mathbf{x^{2}+5\sqrt{3}x+12} =(x+4\sqrt{3})(x+\sqrt{3})

Example 5: Factorise \mathbf{9x^{2}-22x+8}.

Sol. Here , 9 x 8 =72. We spilt -22 into parts whose sum is-22 and product 72 .

Clearly , (-18)+(-4) =-22 and (-18) x (-4)=72 .

\therefore 9x^{2}-22x+8=9x^{2}-18x-4x+8

=9x(x-2)-4(x-2)

= (x-2)(9x-4)

Hence \mathbf{9x^{2}-22x+8}= (x-2)(9x-4)

Example 6: Factorise \mathbf{6\sqrt{3}x^{2}-47x+5\sqrt{3}}.

Sol. Here , 6\sqrt{3}\times 5\sqrt{3}=90. We spilt -47 into two parts whose sum is-47 and product 90 .

Clearly , (-2)+(-45) =-47 and (-2) x (-45)=90 .

6\sqrt{3}x^{2}-47x+5\sqrt{3}=6\sqrt{3}x^{2}-2x-45x+5\sqrt{3

=2x(3\sqrt{3}x-1)-5\sqrt{3}(3\sqrt{3}x-1)

= (2x-5\sqrt{3})(3\sqrt{3}x-1)

Hence \mathbf{6\sqrt{3}x^{2}-47x+5\sqrt{3}}= (2x-5\sqrt{3})(3\sqrt{3}x-1).

Example 7: Factorise \frac{3}{5}x^{2}-\frac{19}{5}x +4 .

Solution : Here, \frac{3}{5} \times 4 = \frac{12}{5}  . We split -\frac{19}{5} into two parts whose sum is -\frac{19}{5} and product \frac{12}{5} .

Clearly, -3-\frac{4}{5}=-\frac{19}{5} and (-3)\times (-\frac{4}{5})=\frac{12}{5}  .

\frac{3}{5}x^{2}-\frac{19}{5}x +4\frac{3}{5}x^{2}-3x-\frac{4}{5}x +4

= 3x(\frac{x}{5}-1)-4(\frac{x}{5}-1)

=(\frac{x}{5}-1)(3x-4)

Hence \frac{3}{5}x^{2}-\frac{19}{5}x +4 = (\frac{x}{5}-1)(3x-4).

Example 8: Factorise \boldsymbol{7\sqrt{2}x^{2}-10x-4\sqrt{2}} .

Solution: The given expression is 7\sqrt{2}x^{2}-10x-4\sqrt{2} .

Here , 7\sqrt{2}\times (-4\sqrt{2})=-56.

So, we split -10  in two parts whose sum is -10 and product -56.

Clearly, (-14+4) =-10 and  (-14) x 4 =-56.

\therefore 7\sqrt{2}x^{2}-10x-4\sqrt{2} =7\sqrt{2}x^{2}-14x+4x-4\sqrt{2}

=7\sqrt{2}x(x- \sqrt{2})+4(x-\sqrt{2})

= (x-\sqrt{2})(7\sqrt{2}x+4)

Factoring trinomials worksheet PDF download

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