Question

Find the value of k for which a-3b is a factor of a4-7a2b2+kb4.Hence, for this value of k , factorise a4-7a2b2+kb4 completely.

Answer 1

The value of k is -18

(a-3b),(a+3b),$(a-bi\sqrt{2})$and $(a+bi\sqrt{2})$ are the factors of the given polynomial

We can also write it as $(a\pm3b)$,$(a\pm bi\sqrt{2})$

The polynomial can be written as $a^4-7a^b^2+kb^4=(a-3b)(a+3b)(a-bi\sqrt{2})(a+bi\sqrt{2})$

Step-by-step explanation:

Given expression is $a^4-7a^b^2+kb^4$ and a-3b is a factor of the given polynomial

Now to find the value of k:

By synthetic division we can solve this polynomial

Since a-3b=0

a=3b is a zero of  $a^4+0a^3b-7a^b^2+0ab^3+kb^4$

3b_|    1       0     $-7b^2$     0         $kb^4$

          0     3b     $9b^2$    $6b^3$      $18b^4$

      __________________________________________________

        1       3b    $2b^2$      $6b^3$    $kb^4+18b^4$

Since a-3b is a factor of given polynomial we have $kb^4+18b^4=0$

$k=\frac{-18b^4}{b^4}$

Therefore the value of k=-18

Now the given polynomial becomes $a^4-7a^b^2-18b^4$

Now factorise the $a^4-7a^b^2-18b^4$ we have

3b_|    1       0     $-7b^2$     0     $-18b^4$

          0     3b     $9b^2$    $6b^3$      $18b^4$

      ______________________________________________

        1       3b    $2b^2$      $6b^3$      0

$a^3+3a^2b+2ab^2+6b^3=0$

Again using synthetic division

-3b_|    1       3b     $2b^2$    $6b^3$          

           0     -3b     0      $-6b^3$  

       _____________________________________

            1       0      $2b^2$      0

Therefore a+3b is a factor

and $a^2+0ab+2b^2=0$

$a^2+2b^2=0$

$a^2=-2b^2$

$a=2b^2i^2$ ( since $i^2=-1$)

$=\pm bi\sqrt{2}$

$a=bi\sqrt{2}$ and $a=-bi\sqrt{2}$ are the zeros

Therefore (a-3b),(a+3b),$(a-bi\sqrt{2})$and $(a+bi\sqrt{2})$ are the factors of the given polynomial

We can also write it as $(a\pm3b)$,$(a\pm bi\sqrt{2})$

The polynomial can be written as $a^4-7a^b^2+kb^4=(a-3b)(a+3b)(a-bi\sqrt{2})(a+bi\sqrt{2})$

Answer 2

Answer:

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