Question

find the value of a and b so that x-1 and x+2 are factors of f(x)=2x³+ax²+bx-14​

Answer 1

x-1=0

x=1

f(1)=2(1)^3+a(1)^2+b(1)-14=0

= 2+a+b-14=0

= a+b-12=0

= a+b=12 _(1)

x+2=0

x=-2

f(-2)=2(-2)^3+a(-2)^2+b(-2)-14=0

= -16+4a-2b-14=0

= -30+4a-2b=0

=4a-2b=30

=2(2a-b)=30

=2a-b=15__(2)

on adding equation (1) and(2),

2a-b=15

a+b=12

b and -b will cancel each other

so,3a=27

a=27/3=9

a=9

put the value of a in equation (1),

a+b=12

9+b=12

b=12-9

b=3

hence, a=9 and b=3

Answer 2

Solution :

$\bf{\large{\underline{\bf{Given\::}}}}}$

We have x - 1 and x + 2 are the factors of f(x) = 2x³ + ax² + bx - 14.

$\bf{\large{\underline{\bf{To\:find\::}}}}}$

The value of a and b.

$\bf{\large{\underline{\bf{Explanation\::}}}}}$

We have one factor of cubic polynomial is x - 1

So;

x - 1 = 0

x = 1

∴Putting the value of x in the given polynomial f(x).

$\longrightarrow\sf{f(x)=2x^{3} +ax^{2} +bx-14=0}\\\\\longrightarrow\sf{f(1)=2(1)^{3} +a(1)^{2} +b(1)-14=0}\\\\\longrightarrow\sf{f(1)=2*1+a*1+b-14=0}\\\\\longrightarrow\sf{f(1)=2+a+b-14=0}\\\\\longrightarrow\sf{f(1)=a+b-12=0}\\\\\longrightarrow\bf{f(1)=a+b=12.....................(1)}$

&

We have other factor of cubic polynomial is x + 2

So;

x + 2 = 0

x = -2

∴Putting the value of x in the given polynomial f(x).

$\longrightarrow\sf{f(x)=2x^{3} +ax^{2} +bx-14=0}\\\\\longrightarrow\sf{f(-2)=2(-2)^{3} +a(-2)^{2} +b(-2)-14=0}\\\\\longrightarrow\sf{f(-2)=2*(-8)+a*4+(-2b)-14=0}\\\\\longrightarrow\sf{f(-2)=-16+4a-2b-14=0}\\\\\longrightarrow\sf{f(-2)=4a-2b-30=0}\\\\\longrightarrow\sf{f(-2)=2(2a-b-15)=0}\\\\\longrightarrow\sf{f(-2)=2a-b-15=0}\\\\\longrightarrow\bf{f(-2)=2a-b=15.........................(2)}$

$\green{\underline{\underline{\bf{Using\:Substitution\:method\::}}}}$

From equation (1),we get;

$\mapsto\sf{a+b=12}\\\\\mapsto\bf{a=12-b............................(3)}$

Putting the value of a in equation (2),we get;

$\mapsto\sf{2(12-b)-b=15}\\\\\mapsto\sf{24-2b-b=15}\\\\\mapsto\sf{24-3b=15}\\\\\mapsto\sf{-3b=15-24}\\\\\mapsto\sf{-3b=-9}\\\\\mapsto\sf{b=\cancel{\dfrac{-9}{-3} }}\\\\\mapsto\sf{\red{b=3}}$

Putting the value of b in equation (3),we get;

$\mapsto\sf{a=12-3}\\\\\mapsto\sf{\red{a=9}}$

Thus;

The value of a is 9 and b is 3 .