How do you completely factor $25 x^{3} + 150 x^{2} + 131 x + 30$ $25x^3+150x^2+131x+30$ given 5x+3 is one of the factors?

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Answer 1 · 2016-10-04T19:10:57

$(5 x + 3) (5 x + 2) (x + 5)$ $(5x+3)(5x+2)(x+5)$ .

Let, $p (x) = 25 x^{3} + 150 x^{2} + 131 x + 30 .$ $p(x)=25x^3+150x^2+131x+30.$

We are given that, $(5 x + 3)$ $(5x+3)$ is a factor of $p (x)$ $p(x)$ .

To factorise $p (x)$ $p(x)$ completely, Synthetic Division is usually

performed, but, we can proceed as under :-

$p (x) = 25 x^{3} + 150 x^{2} + 131 x + 30$ $p(x)=25x^3+150x^2+131x+30$

$=ul(25x^3+15x^2)+ul(135x^2+81x)+ul(50x+30)$

$=5x^2(5x+3)+27x(5x+3)+10(5x+3)$

$=(5x+3)(5x^2+27x+10)$

$=(5x+3){ul(5x^2+25x)+ul(2x+10)}$
.................................................................... $[25xx2=50, 25+2=27]$

$=(5x+3){5x(x+5)+2(x+5)}$

$:. p(x)=(5x+3)(5x+2)(x+5)$ .

How do you completely factor 25x^3+150x^2+131x+3025x3+150x2+131x+3025x^3+150x^2+131x+30 given 5x+3 is one of the factors?