It being given that 1 is …
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Sia ? 6 years, 7 months ago
The given polynomial is -x3 + 7x - 6 and let f (x) = -x3 + 7x - 6 .
Since 1 is a zero of f(x), so (x - 1) is a factor of f(x).
Now we divide f(x) = -x3 + 7x - 6 by (x - 1), we obtain
Where quotient = (-x2 - x + 6)
{tex}\therefore{/tex} f(x) = (-x3 + 7x - 6) = (x - 1)(-x2 - x + 6)
{tex}= - ( x - 1 ) \left( x ^ { 2 } + x - 6 \right) = - ( x - 1 ) \left( x ^ { 2 } + 3 x - 2 x - 6 \right){/tex}
{tex}= ( 1 - x ) [ x ( x + 3 ) - 2 ( x + 3 ) ] = ( 1 - x ) ( x + 3 ) ( x - 2 ){/tex}
{tex}\therefore{/tex} f(x) = 0 {tex}\Rightarrow{/tex} {tex}( 1 - x ) ( x + 3 ) ( x - 2 ){/tex} = 0
{tex}\Rightarrow{/tex} (1 - x ) = 0 or (x + 3) = 0 or (x - 2) = 0
{tex}\Rightarrow{/tex} x = 1 or x = -3 or x = 2.
Thus, the other zeros are -3 and 2
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