Find the values of a and b, if x2 – 4
Question:

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

 

Solution:

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

$g(x)=x^{2}-4$

first we need to find the factors of g(x)

$\Rightarrow x^{2}-4$

$\Rightarrow x^{2}=4$

$\Rightarrow x=\sqrt{4}$

$\Rightarrow x=\pm 2$

(x – 2) and (x + 2) are the factors

By factor therorem if (x – 2) and (x + 2) are the factors of f(x) the result of f(2) and f(-2) should be zero

Let, x – 2 = 0

⟹ x = 2

Substitute the value of x in f(x)

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

= 16a + 2(8) – 3(4) + 2b – 4

= 16a + 2b + 16 – 12 – 4

= 16a + 2b

Equate f(2) to zero

⟹ 16a + 2b = 0

⟹ 2(8a + b) = 0

⟹ 8a + b = 0 ….. 1

Let, x + 2 = 0

⟹ x = -2

Substitute the value of x in f(x)

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

= 16a + 2(-8) – 3(4) – 2b – 4

= 16a – 2b – 16 – 12 – 4

= 16a – 2b – 32

= 16a – 2b – 32

Equate f(2) to zero

⟹ 16a – 2b – 32 = 0

⟹ 2(8a – b) = 32

⟹ 8a – b = 16 …. 2

Solve equation 1 and 2

8a + b = 0

8a – b = 16

16a = 16

a = 1

substitute a value in eq 1

8(1) + b = 0

⟹ b = – 8

The values are a = 1 and b = – 8

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