Answer:
Step-by-step explanation:
Given : ![p(x)=x^{3} +6x^{2} -7x-60](https://tex.z-dn.net/?f=p%28x%29%3Dx%5E%7B3%7D%20%2B6x%5E%7B2%7D%20-7x-60)
Solution :
Part A:
First find the potential roots of p(x) using rational root theorem;
So, ![\text{Possible roots} =\pm\frac{\text{factors of constant term}}{\text{factors of leading coefficient}}](https://tex.z-dn.net/?f=%5Ctext%7BPossible%20roots%7D%20%3D%5Cpm%5Cfrac%7B%5Ctext%7Bfactors%20of%20constant%20term%7D%7D%7B%5Ctext%7Bfactors%20of%20leading%20coefficient%7D%7D)
Since constant term = -60
Leading coefficient = 1
![\text{Possible roots} =\pm\frac{\text{factors of 60}}{\text{factors of 1}}](https://tex.z-dn.net/?f=%5Ctext%7BPossible%20roots%7D%20%3D%5Cpm%5Cfrac%7B%5Ctext%7Bfactors%20of%2060%7D%7D%7B%5Ctext%7Bfactors%20of%201%7D%7D)
![\text{Possible roots} =\pm\frac{1,2,3,4,5,6,10,12,15,20,60}{1}](https://tex.z-dn.net/?f=%5Ctext%7BPossible%20roots%7D%20%3D%5Cpm%5Cfrac%7B1%2C2%2C3%2C4%2C5%2C6%2C10%2C12%2C15%2C20%2C60%7D%7B1%7D)
Thus the possible roots are ![\pm1, \pm2, \pm 3, \pm4, \pm5,\pm6, \pm10, \pm12, \pm15, \pm20, \pm60](https://tex.z-dn.net/?f=%20%5Cpm1%2C%20%5Cpm2%2C%20%5Cpm%203%2C%20%5Cpm4%2C%20%5Cpm5%2C%5Cpm6%2C%20%5Cpm10%2C%20%5Cpm12%2C%20%5Cpm15%2C%20%5Cpm20%2C%20%5Cpm60)
Thus from the given options the correct answers are -10,-5,3,15
Now For Part B we will use synthetic division
Out of the possible roots we will use the root which gives remainder 0 in synthetic division :
Since we can see in the figure With -5 we are getting 0 remainder.
Refer the attached figure
We have completed the table and have obtained the following resulting coefficients: 1 , 1,−12,0. All the coefficients except the last one are the coefficients of the quotient, the last coefficient is the remainder.
Thus the quotient is ![x^{2} +x-12](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2Bx-12)
And remainder is 0 .
So to get the other two factors of the given polynomial we will solve the quotient by middle term splitting
![x^{2} +x-12=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2Bx-12%3D0)
![x^{2} +4x-3x-12=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B4x-3x-12%3D0)
![x(x+4)-3(x+4)=0](https://tex.z-dn.net/?f=x%28x%2B4%29-3%28x%2B4%29%3D0)
![(x-3)(x+4)=0](https://tex.z-dn.net/?f=%28x-3%29%28x%2B4%29%3D0)
Thus x-3 and x+4 are the other two factors
So , p(x)=(x+5)(x-3)(x+4)