Using factor theorem, factorize each of the following polynomials:(i)

1.	Using factor theorem, factorize each of the following polynomials:(i) x3– 6x2+ 3x + 10 (ii) 2y3– 5y – 19y + 42
Answer» $i) Let \: p(x) = x^{3} - 6x^{2}+3x + 10$ $By \:trial ,$ $p(-1) = (-1)^{3} - 6(-1)^{2} + 3(-1) + 10$ $= -1 - 6 - 3 + 10$ $= -10 + 10$ $= 0$ $p(-1) = 0$ $So, (x+1) \: is \:a \:factor \: of \:p(x)$ $Now, divide \: p(x) \: by \: (x+1) .$ 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) $p(y) = (y-2)(2y^{2} - y - 21 )$ $= (y-2)(2y^{2} -7y+6y-21)$ $= (y-2)[ y(2y-7) + 3(2y-7)]$ $= (y-2)(2y-7)(y+3)$ $\therefore \red{2y^{3} -5y^{2} -19y+42}$ $\green {= (y-2)(2y-7)(y+3)}$ •••♪