Answer:
[tex]P(x)=(x+7(x+4)(x-3)(x-5)=x^{4}+3x^{3}-45x^{2}-59x+420[/tex]
Step-by-step explanation:
Given:
The zeros of the polynomial are -7, -4, 3 and 5.
The coefficient of [tex]x^{4}[/tex] is 1.
A polynomial of degree 4 has maximum 4 zeros.
Let the polynomial be [tex]P(x)[/tex].
Since, [tex]P(x)[/tex] has zeros -7, -4, 3 and 5, therefore,
[tex]P(x)=a(x-(-7))(x-(-4))(x-3)(x-5)\\P(x)=a(x+7)(x+4)(x-3)(x-5)[/tex]
Where, [tex]a[/tex] is coefficient of [tex]x^{4}[/tex].
But, coefficient of [tex]x^{4}[/tex] is 1. So, [tex]a=1[/tex]
∴ [tex]P(x)=(x+7)(x+4)(x-3)(x-5)\\P(x)=(x^{2}+11x+28)(x^{2}-8x+15)\\P(x)=x^{4}-8x^{3}+15x^{2}+11x^{3}-88x^{2}+165x+28x^{2}-224x+420\\P(x)=x^{4}+3x^{3}-45x^{2}-59x+420[/tex]
