Question

Using factor theorem, factorize each of the following polynomials:
(i) x3– 6x2+ 3x + 10
(ii) 2y3– 5y – 19y + 42

Answer 1

$i) Let \: p(x) = x^{3} - 6x^{2}+3x + 10$

$By \:trial ,$

$p(-1) = (-1)^{3} - 6(-1)^{2} + 3(-1) + 10$

$= -1 - 6 - 3 + 10$

$= -10 + 10$

$= 0$

$p(-1) = 0$

$So, (x+1) \: is \:a \:factor \: of \:p(x)$

$Now, divide \: p(x) \: by \: (x+1) .$

x+1| x³-6x²+3x+10|x²-7x+10

*****x³+x²

_____________

******* -7x²+3x

*******-7x²-7x

_____________

**********10x+10

********** 10x +10

______________

**********(0)

$\therefore p(x) = (x+1)(x^{2}-7x+10)$

$= (x+1)( x^{2} - 5x - 2x + 10)$

$= (x+1)[ x(x-5) -2(x-5)]$

$= (x+1)(x-5)(x-2)$

$\red{ x^{3} - 6x^{2}+3x + 10}$

$\green { = (x+1)(x-5)(x-2)}$

/* There is a mistake in second problem. It must be like this */

$ii) Let \: p(y) = 2y^{3} -5y^{2} -19y+42$

$By \:trial ,$

$p(2) = 2\times2^{3} - 5\times 2^{2} - 19\times 2 + 42$

$= 16 - 20 - 38 + 42$

$= 58 - 58$

$= 0$

$p(2) = 0$

$\therefore (y-2) \:is \: a \: factor \:of \:p(y)$

$Now, Divide \: p(y) \:by \: (y-2), we \:get$

y-2| 2y³-5y²-19y+42|2y²-y-21

***** 2y³-4y²

____________

********-y²-19y

********-y²+2y

____________

********* -21y+42

********* -21y+42

_____________

***********(0)

$p(y) = (y-2)(2y^{2} - y - 21 )$

$= (y-2)(2y^{2} -7y+6y-21)$

$= (y-2)[ y(2y-7) + 3(2y-7)]$

$= (y-2)(2y-7)(y+3)$

$\therefore \red{2y^{3} -5y^{2} -19y+42}$

$\green {= (y-2)(2y-7)(y+3)}$

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