Is x = 1 a zero of polynomial x3– a2 – x + 1 ?

Let  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>
At straight x equals 1 comma  the value of p(x) is
       straight P left parenthesis 1 right parenthesis space equals space 1 cubed minus 1 squared minus 1 plus 1
space space space space space space space space space equals space 1 minus 1 minus 1 plus 1
space space space space space space space space space equals space 2 space minus 2 space equals 0
Hence, x = 1 is a zero of polynomial x3 – a2 - x + 1.

621 Views

If sum of the squares of zeroes of the quadratic polynomial p(x) = x2 – 8x + K is 40, find the value of K.

We have,
p(x) = x2 – 8x + K
Let α and β are the zeroes of the quadratic polynomial, then
Sum of zeroes left parenthesis straight alpha plus straight beta right parenthesis space equals space fraction numerator negative Coeff. space of space straight x over denominator Coeff. space of space straight x squared end fraction

                                    equals fraction numerator negative left parenthesis negative 8 right parenthesis over denominator 1 end fraction equals 8
And, 
Product of zeroes
                          left parenthesis straight alpha comma space straight beta right parenthesis space equals space fraction numerator Constant space term over denominator Coeff. space of space straight x squared end fraction
space space space space space space space space space space space space equals space straight K over 1 space equals space straight K
We have, 
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#6 {main}</pre>
rightwards double arrow            40 space space equals space left parenthesis 8 right parenthesis squared minus 2 left parenthesis straight K right parenthesis
rightwards double arrow             40 space equals space 64 minus 2 straight K
rightwards double arrow             space space space space space space space space space 2 straight K space equals space 24 space space rightwards double arrow space space straight K space equals 12
Hence,   K = 12.
1773 Views

Find the sum and product of the zeroes of the quadratic polynomial:
p(x) = abx2 + (b2 – ac) x – bc.

We have,
p(x) = abx2 + (b2 – ac) x – bc
Let α and β be the zeroes of the quadratic polynomial p(x), then
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#6 {main}</pre>
                                 equals fraction numerator negative left parenthesis straight b squared minus ac right parenthesis over denominator ab end fraction equals fraction numerator ac minus straight b squared over denominator ab end fraction
And product of the zeroes left parenthesis straight alpha comma space straight beta right parenthesis
                                equals space fraction numerator Constant space term over denominator Coeff. space of space straight x squared end fraction

                                 equals fraction numerator negative bc over denominator ab end fraction equals negative straight c over straight a
98 Views

Find the zeroes of the quadratic polynomial x2 – 5x + 6, and verify the relationship between the zeroes and the coefficients.

We have x2 – 5x + 6 – x2 – 3x – 2x + 6
= (x – 3) (x – 2)
So, the value of x2 – 5x + 6 is zero when x - 3 = 0 or x – 2 = 0
i.e., when x = 3 or x = 2
Therefore, the zeroes of x2 – 5x + 6 are 3 and 2.
Sum of zeroes = 3 plus 2 space equals space 5 over 1
                        = fraction numerator negative left parenthesis Coefficient space of space straight x right parenthesis over denominator Coefficient space of space straight x squared end fraction
Product of zeroes
equals 3 cross times 2 equals 6 over 1 equals fraction numerator Constant space term over denominator Coefficient space of space straight x squared end fraction
512 Views

Check whether the polynomial x2 + 4x + 5 has zeroes. If yes, find zeroes of the polynomial.

Solution not provided.


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If α, β are the zeroes of the polynomial p(x) = 4x2 + 3x + 7. Find the value of space space 1 over straight alpha plus 1 over straight beta.

We have,
p(x) = 4x2 + 3x + 7, then
Sum of the zeroes (α + β)
equals fraction numerator negative Coeff. of space straight x over denominator Coeff. space of space straight x squared end fraction equals fraction numerator negative 3 over denominator 4 end fraction
And, product of the zeroes (α. β)
                                equals space fraction numerator Constant space term over denominator Coeff. of space straight x squared end fraction equals 7 over 4

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#6 {main}</pre>

                              equals fraction numerator negative begin display style 3 over 4 end style over denominator begin display style 7 over 4 end style end fraction equals negative 3 over 7. Ans.  
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