Bush Bob
05/20/2023 · High School
\begin{tabular}{|l|l} Factorize: \\ \( \begin{array}{lll}\text { (a) } a^{2}+12 a+36 & \text { (b) } y^{2}+14 y+49 & \text { (c) } p^{2}+22 p+121 \\ \text { (d) } 4 a^{2}+20 a+25 & \text { (e) } 9 \mathrm{r}^{2}+60 \mathrm{r}+100 & \text { (f) } 36 x^{2}+84 x+49 \\ \text { (g) } x^{2}-8 x+16 & \text { (h) } a^{2}-18 a+81 & \text { (i) } p^{2}-26 p+169\end{array} \)\end{tabular}
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To factor the given expressions, we will look for perfect square trinomials and apply the factoring formulas. Here are the factorizations for each part:
(a) \( a^{2}+12a+36 \)
This is a perfect square trinomial:
\[
a^{2}+12a+36 = (a+6)^{2}
\]
(b) \( y^{2}+14y+49 \)
This is also a perfect square trinomial:
\[
y^{2}+14y+49 = (y+7)^{2}
\]
(c) \( p^{2}+22p+121 \)
This is another perfect square trinomial:
\[
p^{2}+22p+121 = (p+11)^{2}
\]
(d) \( 4a^{2}+20a+25 \)
We can factor out a perfect square:
\[
4a^{2}+20a+25 = (2a+5)^{2}
\]
(e) \( 9r^{2}+60r+100 \)
This is also a perfect square trinomial:
\[
9r^{2}+60r+100 = (3r+10)^{2}
\]
(f) \( 36x^{2}+84x+49 \)
This is another perfect square trinomial:
\[
36x^{2}+84x+49 = (6x+7)^{2}
\]
(g) \( x^{2}-8x+16 \)
This is a perfect square trinomial:
\[
x^{2}-8x+16 = (x-4)^{2}
\]
(h) \( a^{2}-18a+81 \)
This is a perfect square trinomial:
\[
a^{2}-18a+81 = (a-9)^{2}
\]
(i) \( p^{2}-26p+169 \)
This is a perfect square trinomial:
\[
p^{2}-26p+169 = (p-13)^{2}
\]
In summary, the factorizations are:
- (a) \( (a+6)^{2} \)
- (b) \( (y+7)^{2} \)
- (c) \( (p+11)^{2} \)
- (d) \( (2a+5)^{2} \)
- (e) \( (3r+10)^{2} \)
- (f) \( (6x+7)^{2} \)
- (g) \( (x-4)^{2} \)
- (h) \( (a-9)^{2} \)
- (i) \( (p-13)^{2} \)
Quick Answer
- (a) \( (a+6)^{2} \)
- (b) \( (y+7)^{2} \)
- (c) \( (p+11)^{2} \)
- (d) \( (2a+5)^{2} \)
- (e) \( (3r+10)^{2} \)
- (f) \( (6x+7)^{2} \)
- (g) \( (x-4)^{2} \)
- (h) \( (a-9)^{2} \)
- (i) \( (p-13)^{2} \)
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