Question

Find the values of a and b, if $x^2$ - 4 is a factor of $a{x}^4+2x^3-3x^2$ + bx - 4

Answer 1

f(x) = $a{x}^4+2x^3-3x^2$ + bx - 4

Factors of $x^2-4=(x{)}^2-(2{)}^2$

= (x+2)(x-2)

If x + 2 = 0, then x = - 2

Now, f(-2) = $a(-2{)}^4+2(-2{)}^3-3(-2{)}^2+b(-2)-4$

16a - 16 - 12 -2b - 4

= 16a - 2b - 32

$\because$ x + 2 is a factor of f(x)

$\therefore$ Remainder = 0

$\Rightarrow$ 16a - 2b - 32 = 0

$\Rightarrow$ 8a - b - 16 = 0 $\Rightarrow$ 8a - b = 16 .......(i)

Again x - 2 = 0, then x = 2

Now f(2) = a $\times (2{)}^4+2(2{)}^3-3(2{)}^2+b\times$ 2-4

= 16a + 2b

$\because$ x - 2 is a factor of f(x)

$\therefore$ Remainder = 0

$\Rightarrow$ 16a + 2b = 0 $\Rightarrow$ 8a + b = 0 .......(ii)

Adding (i) and (ii),

16a = 16 $\Rightarrow a=\frac{16}{16}=1$

From (ii) 8$\times$ 1 + b = 0 $\Rightarrow$ 8 + b = 0

$\Rightarrow$ b = - 8

$\therefore$ a = 1, b = - 8