Question

Find all the zeros of the polynomial x 4 - 3 x cube minus 5 x square + 21 x minus 14 if two of its zeros are root 7 and minus root 7

Answer 1

$\textbf{\huge{ Hello Mate! }}$

Please refer to attatchment.

Now, further solution.

We already had two zeros whom when we multiplied we got $x^2 - 7$

Now, since dividend = quotient × divisor.

Here, dividend is assumed p(x)

$p(x) = (x - \sqrt{7} )(x - \sqrt{7} )( {x}^{2} - 3x + 2) \\ \\ p(x) = (x - \sqrt{7} )(x + \sqrt{7} )( {x}^{2} - 2x - x + 2) \\ \\ p(x) = (x - \sqrt{7} )(x + \sqrt{7} )(x(x - 2) - 1(x - 2)) \\ \\ p(x) = (x - \sqrt{7} )(x + \sqrt{7} )(x - 1)(x + 2)$

Hence, zeros or value of x = $\sqrt{7} , - \sqrt{7} , 1 and - 2$

$\textbf{ Have great future ahead! }$

Answer 2

here is your answer OK


Since 7 and -7 are the zeroes of the given polynomial.
This means x-7 and x+7 are the factors of the given polynomial.
So dividing px = x4-3x3-5x2+21x-14 by x-7x+7 i.e. x2-7 using long division method we get;
x2-3x+2x2-7)x4-3x3-5x2+21x-14 x4 -7x2 -3x3+2x2+21x -3x3 +21x 2x2 -14 2x2-14 0
So x2-3x+2 is the factor of the given polynomial. To find other zeroes equate it to 0 we get;
x2-3x+2 = 0⇒x2-2x-x+2 = 0⇒xx-2-1x-2 = 0⇒x-1x-2 = 0⇒x = 1 and x = 2
Therefore other two zeroes of the given polynomial are 1 and 2.