Practice Factoring Trinomials

1. Factor ‍$$x^2+7x+10$$.

   Choose 1 answer:
   

   1. $$(x-2)(x+5)$$
   2. $$(x+7)(x+10)$$

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

2. Factor ‍$$x^2+9x+20$$.

3. Factor ‍$$x^2-8x-9$$.

4. Factor ‍$$x^2-10x+24$$.

5. Factor ‍$$x^2+7x-30$$.

6. Can this factorization method be used to factor ‍$$2x^2+3x+1$$.

   Choose 1 answer:
   

   1. Yes

   2. No

7. Factor ‍$$x^2+5xy+6y^2$$.

8. Factor ‍$$x^4-5x^2+6$$.

Source: Khan Academy, https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:factor-quadratics-intro/a/factoring-quadratics-leading-coefficient-1
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1. To factor ‍$$x^2+7x+{10}$$, we need to find factors of ‍ 10 that add up to 7.

   The table below shows all possible factors of 10 and their sums.

   Product $$m\cdot n=10$$	Sum: $$m+n$$
   $$1\cdot 10=10$$	$$1+10=11$$
   $${2\cdot 5=10}$$	$$2+5=7$$
   $$(-1)\cdot (-10)=10$$	$$(-1)+(-10)=-11$$
   $$(-2)\cdot (-5)=10$$	$$(-2)+(-5)=-7$$

   
   

   Only one pair of factors satisfies the conditions of the problem: ‍2 and 5.

   We can add each of these numbers to $$x$$ to form the two binomial factors: ‍$$(x+2)$$ and ‍$$(x+5)$$. The factorization of the polynomial is given below.

   $$x^2+7x+10=(x+2)(x+5)$$

2. To factor ‍$$x^2+9x+{20}$$, we need to find factors of ‍ 20 that add up to 9.

   The table below shows all possible factors of 20 and their sums.

   Product $$m\cdot n=20$$	Sum: $$m+n$$
   $$1\cdot 20=20$$	$$1+20=21$$
   $${2\cdot 10=20}$$	$$2+10=12$$
   $$(4)\cdot (5)=20$$	$$4+5=9$$
   $$(-1)\cdot (-20)=20$$	$$(-1)+(-20)=-21$$
   $$(-2)\cdot (-10)=20$$	$$(-2)+(-10)=-12$$
   $$(-4)\cdot (-5)=20$$	$$(-4)+(-5)=-9$$

   
   

   Only one pair of factors satisfies the conditions of the problem: ‍4 and 5.

   We can add each of these numbers to $$x$$ to form the two binomial factors: ‍$$(x+4)$$ and ‍$$(x+5)$$. The factorization of the polynomial is given below.

   $$x^2+9x+20=(x+4)(x+5)$$

3. To factor ‍$$x^2-8x-9$$, we need to find factors of ‍-9 that add up to -8.

   Since the two numbers must multiply to -9, one must be positive and one must be negative.
   

   The factors of ‍-9 are listed in the table below.
   

   Product $$m\cdot n=-9$$	Sum: $$m+n$$
   $$1\cdot ({-9})=-9$$	$$1+({-9})=-8$$
   $$(-1)\cdot 9=-9$$	$$-1+9=8$$
   $$(-3)\cdot 3=-9$$	$$-3+3=0$$

   
   

   Notice here that ‍1 and ‍-9 satisfy these requirements. We can add each of these numbers to ‍$$x$$ to form the two binomial factors: ‍$$(x+1)$$ and ‍$$(x+(-9))$$. The factorization of the polynomial is given below.

   $$ x^2-8x-9 =(x+1)(x+({-9})) \\ \quad =(x+1)(x-9) $$

4. To factor ‍$$x^2-10x+24$$, we need to find factors of ‍24 that add up to -8.

   Since the two numbers must multiply to 24 and add to -10, both numbers must be negative.
   

   The factors of ‍24 (where both are negative) are listed in the table below.
   

   Product $$m\cdot n=24$$	Sum: $$m+n$$
   $$(-1)\cdot ({-24})=24$$	$$-1+({-24})=-25$$
   $$(-2)\cdot (-12)=24$$	$$-2+(-12)=-14$$
   $$(-3)\cdot (-8)=24$$	$$-3+(-8)=-11$$
   $$({-4})\cdot ({-6})=24$$	$${-4}+({-6})=-10$$

   
   

   Notice here that ‍-4 and ‍-6 satisfy these requirements. We can add each of these numbers to ‍$$x$$ to form the two binomial factors: ‍$$(x+(-4))$$ and ‍$$(x+(-6))$$. The factorization of the polynomial is given below.

   $$ x^2-10x-9 =(x+(-4))(x+({-6})) \\\quad =(x-4)(x-6) $$

5. $$=(x-3)(x+10)$$

6. No

   In this method, we factor the trinomial by equating it to ‍$$(x+m)(x+n)$$ for some integers ‍$$m$$ and ‍$$n$$. This means that the leading term of this polynomial will always be ‍$$x^2$$. Since the leading term of this polynomial is ‍$$2x^2$$, it cannot be factored using this method.

7. Recall the following factorization:

   $$x^2+5x+6=(x+2)(x+3)$$

   The polynomial ‍$$x^2+5xy+6y^2$$ is very similar to ‍$$x^2+5x+6$$. The only difference is that the middle term is multiplied by ‍$$y$$ and the last term is multiplied by ‍$$y^2$$.

   We can obtain this by adding a factor of ‍$$y$$ to each binomial:

   $$\begin{aligned}(x+2y)(x+3y)
   &=x^2+3xy +2x y+6{y^2}\\
   \\
   &=x^2+5xy+6{y^2}
   \end{aligned}$$
   

   The factorization is $$(x+2y)(x+3y)$$

8. This polynomial is of the form ‍$$x^2+bx+c$$, although a substitution makes this easier to see.

   Let $$Y=x^2$$. We can now write the polynomial as follows:

   $$\begin{aligned}x^4-5x^2+6&=({x^2})^2-5{x^2}+6\\
   \\
   &={Y}^2-5{Y}+6\\
   \end{aligned}$$
   

   Since $$(-2)\cdot (-3)=6$$ and $$(-2)+(-3)=-5$$, the polynomial factors as follows:

   $$=(Y-2)(Y-3)$$
   

   Now, since $$Y=x^2$$, we can back substitute to find a factorization of the original polynomial.

   $$(x^2-2)(x^2-3)$$
   

   The factorization is $$(x^2-2)(x^2-3)$$.