Question

Write the polynomial in factored form. p(x)=(x+5)(x-___)(x+___)

Answer 1

P(x) = (x + 5)(x - 3)(x + 4)

Answer 2

Answer:

Step-by-step explanation:

Given : $p(x)=x^{3} +6x^{2} -7x-60$

Solution :

Part A:

First find the potential roots of p(x) using rational root theorem;

So, $\text{Possible roots} =\pm\frac{\text{factors of constant term}}{\text{factors of leading coefficient}}$

Since constant term = -60

Leading coefficient = 1

$\text{Possible roots} =\pm\frac{\text{factors of 60}}{\text{factors of 1}}$

$\text{Possible roots} =\pm\frac{1,2,3,4,5,6,10,12,15,20,60}{1}$

Thus the possible roots are  $\pm1, \pm2, \pm 3, \pm4, \pm5,\pm6, \pm10, \pm12, \pm15, \pm20, \pm60$

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 $x^{2} +x-12$

And remainder is 0 .

So to get the other two factors of the given polynomial we will solve the quotient by middle term splitting

$x^{2} +x-12=0$

$x^{2} +4x-3x-12=0$

$x(x+4)-3(x+4)=0$

$(x-3)(x+4)=0$

Thus x-3 and x+4 are the other two factors

So , p(x)=(x+5)(x-3)(x+4)