Answer:
[tex](a) 28x^2-23x-15[/tex]
[tex](b) (7x^4-4y^{7})(7x^4+4y^{7})[/tex]
[tex](c)(7x+2)(x+2)[/tex]
Step-by-step explanation:
1. (7x+3)(4x−5)
First, Distribute
(7x+3)(4x−5) = 7x(4x-5)+3(4x-5)
Next, we expand the brackets
[tex]7x(4x-5)+3(4x-5)=28x^2-35x+12x-15[/tex]
Simplifying like terms
[tex](7x+3)(4x-5)=28x^2-23x-15[/tex]
2. [tex]49x^8-16y^{14}[/tex]
Looking at the terms, you notice that each of the numbers is a perfect square.
[tex]49x^8-16y^{14}=7^{2}x^8-4^2y^{14}[/tex]
Next, we express the variable terms as a power of 2. This is to enable us apply a particular principle in factorization.
[tex]7^{2}(x^4)^2-4^2(y^{7})^2= (7x^4)^{2}-(4y^{7})^2[/tex]
Now, we apply the Principle of Difference of Two Squares.
[tex]a^2-b^2=[/tex](a-b)(a+b)
Therefore:
[tex](7x^4)^{2}-(4y^{7})^2=(7x^4-4y^{7})(7x^4+4y^{7})[/tex]
3. [tex]7x^2+16x+4[/tex]
[tex]7x^2+16x+4 = 7x^2+14x+2x+4[/tex]
[tex]=7x(x+2)+2(x+2)\\ =(7x+2)(x+2)[/tex]
