Question

Find the value of A and B so that polynomial (x4+ax3-7x2+8x+b) is exactly divisible by(x+2) as well as (x+3).

Answer 1

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Answer 2

The value of a = $\frac{22}{19}$  and the value of b = $\frac{708}{19}$ .

Step-by-step explanation:

We are given that polynomial (x4+ax3-7x2+8x+b) is exactly divisible by (x+2) as well as (x+3).

The two divisors in the question are (x + 2) and (x + 3). As it is given that the given polynomial is divisible by both these divisors, that means;

x + 2 = 0   and   x + 3 = 0

x = -2  and  x = -3 will make the remainder zero when these values of x are substituted in the given polynomial.

f(x) = $x^{4}+ax^{3} -7x^{2} +8x+b$. So, f(-2) and f(-3) will be equal to zero.

f(-2) = $(-2)^{4}+a(-2)^{3} -7(-2)^{2} +8(-2)+b =0$

f(-3) = $(-3)^{4}+a(-3)^{3} -7(-3)^{2} +8(-3)+b =0$

f(-2) = $16-8a -28 -16+b =0$

       = $-8a+b -28 =0$  ---------------- [Equation 1]

f(-3) = $81-27a -63 -24+b =0$

      = $-27a+b -6 =0$   ---------------- [Equation 2]

Now using the elimination method to find the values of a and b;

                    $-8a+b -28 =0$

                    $-27a+b -6 =0$

                    +        -     +       -  

                         19a - 22 = 0

                            a = $\frac{22}{19}$

Putting the value of a in equation 1 we get;

                          $-8a+b -28 =0$

                          $-8(\frac{22}{19}) +b -28 =0$

                             b  =  $28 + \frac{176}{19}$

                             b  =  $\frac{708}{19}$ .

Hence, the value of a = $\frac{22}{19}$  and the value of b = $\frac{708}{19}$ .