Question

obtain all zeroes of f(x)=xcube+13xsquare+32x+20​

Answer 1

$\huge\mathcal{Given}\\f(x)=x^3+13x^2+32x+20\\\pm1,\pm2,\pm4,\pm5,\pm10\:\:be\:a \:root\:for\:the\: above\:polynomial\\By\:verifacation\:we\:get\\f(-1)=-1^3+13(-1)^2+32(-1)+20\\\:\:\:=-1+13-32+20=-33+33=0\\So,\:(-1)be\:a\:zero\:for\:the\:above\:polynomial \\Now,\:apply\:synthetic\:division\\\:\:\:\:-1\:\:\:\:1\:\:\:\:\:\:13\:\:\:\:\:\:\:\:32\:\:\:\:\:\:\:\:20\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0\:\:\:-1\:\:\:-12\:\:\:-20\\\:\:\:\:\:\:\:\:\:\:\:\:\rule{3cm}{0.5pt}\\\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:1\:\:\:\:\:\:12\:\:\:\:\:\:20\:\:\:\:\:\:0\\So,\:the\:new\:equation\:be\\f(x)=x^2+12x+20</p><p>\\=x^2+10x+2x+20</p><p>\\=x(x+10)+2(x+10)</p><p>\\=(x+10).(x+2)</p><p>\\=>x=-\:10and-\:2\\\huge\mathcal{The\:final\:answer\:is}\\\huge\mathcal{\:x=-1,-2,-10.}</p><p>$

Hope this helps, if you have any doubts lemme know :))))