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Find the value of $a$, if $x-2$ is a factor of $2 x^{5}-6 x^{4}-2 a x^{3}+6 a x^{2}+4 a x+8$.
Find the value of $a$, if $x-2$ is a factor of $2 x^{5}-6 x^{4}-2 a x^{3}+6 a x^{2}+4 a x+8$.
Class 10 | Exercise 8A | ICSE | Maths | Remainder and Factor Theorems | Selina
Find the value of $a$, if $x-2$ is a factor of $2 x^{5}-6 x^{4}-2 a x^{3}+6 a x^{2}+4 a x+8$.

Given, f(x) = 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8 and x – 2 is a factor of f(x).

We know that when a polynomial f (x) is divided by (x – a), the remaining is f from the remainder theorem (a).

So, x – 2 = 0; x = 2

Hence, f(2) = 0

2(2)5 – 6(2)4 – 2a(2)3 + 6a(2)2 + 4a(2) + 8 = 0

64 – 96 – 16a + 24a + 8a + 8 = 0

-24 + 16a = 0

16a = 24

Thus, value of a = 1.5

i

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