Question

Factorise 2x⁴+x³-14x²-19x-6

Answer 1

2x⁴+x³-14x²-19x-6

Putting x=3

p(3)=2(3)⁴+(3)³-14(3)²-19(3)-6

=>0

(x-3) is a factor

__________________

Dividing 2x⁴+x³-14x²-19x-6 by x-3

x-3) 2x⁴+x³-14x²-19x-6( 2x³+7x²+7x+2

-2x⁴+6x³

______

7x³-14x²

-7x³+21x²

_______

7x²-19x

-7x²+21x

_______

2x-6

-2x+6

_______

00

=>2x³+7x²+7x+2

≈≈≈×≈≈≈≈≈×

2x⁴+x³-14x²-19x-6

=>(x-3)(2x³+7x²+7x+2)

=>(x-3){2(x³+1)+7x(x+1)}

=>(x-3)(2+7x)(x³+x+2)

Answer 2

$\huge\tt\green{\underline{\underline{Solution:}}}$

$\sf\underline{Let\:the\:given\:polynomial\:be\:denoted\:by\:p(x),\:then}$

$\sf{p(x)}$=$\sf{2x^4+x^3-14x^2-19x-6}$

Since -6 is the constant term in p(x), therefore , the factors are $\underline{+}$1, $\underline{+}$2, $\underline{+}$3, $\underline{+}<strong>6</strong>$

★$\sf\underline{If\:we\:substitute\:x=-2\:in\:p(x),\:then}$

$\sf{p(-2)=\:2(-2)^4+(-2)^3-14(-2)^2-19(-2)-6}$

$\sf{=32-8-56+38-6}$

$\sf{=(32+38)+(-8-56-6)}$

$\sf{=70-70}$

$\sf{=0}$

$\implies$$\sf{(x+2)\:is\:a\:factor\:of\:p(x)}$

★$\sf\underline{If\:we\:substitute\:x=-3\:in\:p(x),\:then}$

$\sf{p(3)=\:2(3)^4+(3)^3-14(3)^2-19(3)-6}$

$\sf{=162+27-14(9)-57-6}$

$\sf{=189-189}$

$\sf{=0}$

$\implies$$\sf{(x-3)\:is\:a\:factor\:of\:p(x)}$

As (x+2) and (x-3) are factors of p(x), therefore g(x) = (x+2)(x-3) = x² - x - 6 is a factor of p(x). Let divide p(x) by g (x)to obtain the quotient as a quadratic in x and then factorise the quotient.

So , when you divide p(x) by g(x) you will get 2x² + 3x + 1.

•°• 2x⁴ + x³ - 14x² - 19x - 6 = (x² -x- 6)(2x⁴ + 3x + 1)

$\sf{(x+2)(x-3)[2x^2+3x+x+1]}$

$\sf{(x+2)(x-3)[2x(x+1)+1(x+1)]}$

$\sf\underline{(x+2)(x-3)(x+1)(2x+1)}$