Answer:
![P(x)=(x+7(x+4)(x-3)(x-5)=x^(4)+3x^(3)-45x^(2)-59x+420](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8w6zvlujt57i2ltwuv8he9l1wut8n54zxj.png)
Explanation:
Given:
The zeros of the polynomial are -7, -4, 3 and 5.
The coefficient of
is 1.
A polynomial of degree 4 has maximum 4 zeros.
Let the polynomial be
.
Since,
has zeros -7, -4, 3 and 5, therefore,
![P(x)=a(x-(-7))(x-(-4))(x-3)(x-5)\\P(x)=a(x+7)(x+4)(x-3)(x-5)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lto8tzts4dvpk8xvzsq5ytje5u2111j9od.png)
Where,
is coefficient of
.
But, coefficient of
is 1. So,
![a=1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/krkbqffkhyhi4myj6mrpvfi4x1hu9srce5.png)
∴
![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](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xybmuqp6bgglde4ygn52lc6y6y3rnv546o.png)