Question

factoring special products
example 3 factor (a) $x^{2}-4$ and (b) $x^{2}-14x + 49$.
a. $x^{2}-4=x^{2}-2^{2}=(x + 2)(x - 2)$
so, $x^{2}-4=(x + 2)(x - 2)$.
b. $x^{2}-14x + 49=x^{2}-2(x)(7)+7^{2}=(x - 7)^{2}$
so, $x^{2}-14x + 49=(x - 7)^{2}$.
factor the polynomial.

1. $x^{2}-36$
2. $x^{2}-9$
3. $4x^{2}-25$
4. $x^{2}-22x + 121$
5. $x^{2}+28x + 196$
6. $49x^{2}+210x + 225$
7. structure determine the possible integer values of $a$ and $c$ for which the trinomial $ax^{2}+8x + c$ is factorable using the perfect - square trinomial pattern. explain your reasoning.

Explanation:

Step1: Recall difference - of - squares formula

The difference - of - squares formula is $a^{2}-b^{2}=(a + b)(a - b)$.

Step2: Factor $x^{2}-36$

We can write $x^{2}-36$ as $x^{2}-6^{2}$. Using the difference - of - squares formula with $a = x$ and $b = 6$, we get $(x + 6)(x - 6)$.

Step3: Factor $x^{2}-9$

Write $x^{2}-9$ as $x^{2}-3^{2}$. Using the difference - of - squares formula with $a=x$ and $b = 3$, we have $(x + 3)(x - 3)$.

Step4: Factor $4x^{2}-25$

Rewrite $4x^{2}-25$ as $(2x)^{2}-5^{2}$. Using the difference - of - squares formula with $a = 2x$ and $b = 5$, we obtain $(2x+5)(2x - 5)$.

Step5: Recall perfect - square trinomial formula

The perfect - square trinomial formula is $a^{2}-2ab + b^{2}=(a - b)^{2}$ and $a^{2}+2ab + b^{2}=(a + b)^{2}$.

Step6: Factor $x^{2}-22x + 121$

We can write $x^{2}-22x + 121$ as $x^{2}-2(x)(11)+11^{2}$. Using the perfect - square trinomial formula with $a = x$ and $b = 11$, we get $(x - 11)^{2}$.

Step7: Factor $x^{2}+28x + 196$

Write $x^{2}+28x + 196$ as $x^{2}+2(x)(14)+14^{2}$. Using the perfect - square trinomial formula with $a = x$ and $b = 14$, we have $(x + 14)^{2}$.

Step8: Factor $49x^{2}+210x + 225$

Rewrite $49x^{2}+210x + 225$ as $(7x)^{2}+2(7x)(15)+15^{2}$. Using the perfect - square trinomial formula with $a = 7x$ and $b = 15$, we obtain $(7x + 15)^{2}$.

Step9: Solve for $a$ and $c$ in $ax^{2}+8x + c$

For the trinomial $ax^{2}+8x + c$ to be a perfect - square trinomial, it must be in the form $(mx + n)^{2}=m^{2}x^{2}+2mnx + n^{2}$. So, $2mn=8$, which means $mn = 4$.
If $m = 1$ and $n = 4$, then $a=m^{2}=1$ and $c=n^{2}=16$.
If $m = 2$ and $n = 2$, then $a=m^{2}=4$ and $c=n^{2}=4$.
If $m = 4$ and $n = 1$, then $a=m^{2}=16$ and $c=n^{2}=1$.

Answer:

1. $(x + 6)(x - 6)$
2. $(x + 3)(x - 3)$
3. $(2x+5)(2x - 5)$
4. $(x - 11)^{2}$
5. $(x + 14)^{2}$
6. $(7x + 15)^{2}$
7. $a = 1,c = 16$; $a = 4,c = 4$; $a = 16,c = 1$