Question

plz give solution of above question...


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Answer 1

$\huge \mathfrak \red{ÀÑẞWÉR}$

Step-by-step explanation:

$sinc \: {x}^{2} + 2x + kis \: a \: factor \: of \: 2x ^{4} + x ^{3} - 14 {x}^{2} + 5x + 6$

$so \: we \: dividing \: 2 {x}^{4} + x^{3} - 14 {x}^{2} + 5x + 6 \: by \: {x}^{2} + 2x + k \: should \: leave \: tha \: remainder \: 0as \: show \: below$

$Comparing \: the \: coefficient \: of \: x, \: we \: get$

$21+7k=0$

$⇒7k=−21$

$⇒k=− \frac{21}{7} =−3$

$The \: equation  {x}^{2} +2x+k  becomes  {x}^{2} +2x−3 \: and \: the \: factors \: of   {x}^{2} +2x−3=0 \:  are:$

${x}^{2} +3x−x−3=0$

$x(x+3)−1(x+3)=0$

$(x−1)(x+3)=0$

$x=1,−3$

$Now, \: the \: equation 2 {x}^{2} −3x−(8+2k)  \: becomes 2 {x}^{2} −3x−2 and \: the \: factors \: of  2 {x}^{2} −3x−2=0  \: are:$

$2 {x}^{2} −4x+x−2=0$

$2x(x−2)+1(x−2)=0$

$(2x+1)(x−2)=0$

$x=− \frac{1}{2}.2$

$Hence, \: the \: zeros \: of \: the \: given \: two \: polynomials \: are 1,−3,− \frac{1}{2}  and  \: 2.$

$\pink{i \: hope \: it \: helpfull \: for \: you}$

Answer 2

Answer:

Im fine dear

Im in 12th grade

what about you..