Answer:
a) [tex](7\cdot m^{6}+4)\cdot (7\cdot m^{6}-4)[/tex]
b) [tex](7\cdot a\cdot b^{8}+1)\cdot (7\cdot a\cdot b^{8}-1)[/tex]
c) [tex]a = 0.7\cdot t^{9}[/tex]
d) There are two possible answers only considering real coefficients:
(i) [tex]16\cdot x^{2}-81 = (4\cdot x +9)\cdot (4\cdot x - 9)[/tex]
(ii) [tex]14\cdot x^{2}-81 =(\sqrt{14}\cdot x+9)\cdot (\sqrt{14}\cdot x - 9)[/tex]
e) [tex](a^{3}+4)\cdot (a^{3}-4)[/tex]
Step-by-step explanation:
Now we proceed to solve each algebraic equation:
a) Factor [tex]49\cdot m^{12}-16[/tex]
This binomial is of the form [tex]a^{2}-b^{2} = (a+b)\cdot (a-b)[/tex]. In this case, we can rewrite and factor the equation below:
[tex]49\cdot m ^{12}-16[/tex]
[tex](7\cdot m^{6})^ 2-4^{2}[/tex]
[tex](7\cdot m^{6}+4)\cdot (7\cdot m^{6}-4)[/tex]
b) Factor [tex]49\cdot a^{2}\cdot b^{16}-1[/tex]
This binomial is of the form [tex]a^{2}-b^{2} = (a+b)\cdot (a-b)[/tex]. In this case, we can rewrite and factor the equation below:
[tex]49\cdot a^{2}\cdot b^{16}-1[/tex]
[tex](7\cdot a\cdot b^{8})^{2}-1^{2}[/tex]
[tex](7\cdot a\cdot b^{8}+1)\cdot (7\cdot a\cdot b^{8}-1)[/tex]
c) Nicole is factoring [tex]0.49\cdot t^{18}-25[/tex] by using the rule [tex]a^{2}-b^{2} = (a+b)\cdot (a-b)[/tex]. What will she use for the value of [tex]a[/tex]?
From this rule we find that [tex]a^{2} = 0.49\cdot t^{18}[/tex], that is:
[tex]a^{2} = 0.49\cdot t^{18}[/tex]
[tex]a^{2} = \frac{49}{100}\cdot t^{18}[/tex]
[tex]a = \frac{7}{10}\cdot t^{9}[/tex]
[tex]a = 0.7\cdot t^{9}[/tex]
d) Which binomial is a difference of squares?
A binomial is a difference of square if and only if [tex]a^{2}-b^{2} = (a+b)\cdot (a-b)[/tex]. There are two possible answers only considering real coefficients:
(i) [tex]16\cdot x^{2}-81 = (4\cdot x +9)\cdot (4\cdot x - 9)[/tex]
(ii) [tex]14\cdot x^{2}-81 =(\sqrt{14}\cdot x+9)\cdot (\sqrt{14}\cdot x - 9)[/tex]
e) Factor [tex]a^{6}-16[/tex]
This binomial is of the form [tex]a^{2}-b^{2} = (a+b)\cdot (a-b)[/tex]. In this case, we can rewrite and factor the equation below:
[tex]a^{6}-16[/tex]
[tex](a^{3})^{2}-4^{2}[/tex]
[tex](a^{3}+4)\cdot (a^{3}-4)[/tex]
