Question

please solve this question​

Answer 1

★Solution★

Let, p(x)=0 [as, x is a root of p(x)]

p(x) = 2x²–8x+6=0

p(x) = 2(x²–4x+3)=0

[Taking 2 as common]

p(x)=x²–4x+3=0

[Dividing by 2 on both sides]

We, get;

p(x)=x²–4x+3=0

Here, Sum = – 4

Product = 3

Factors =±1, ±3

Pair of such numbers = –3, –1

• So, p(x) can be written as:

p(x)= x²–3x–x+3=0

» x(x–3)–1(x–3)=0

[Taking x and –1 as common]

» (x–1)(x–3)=0

[Taking (x–3) as common]

Thus, either, (x–1)=0 or (x–3)=0

» x=1 or x=3

♦Answer♦

Therefore, the roots of p(x) are 'x'=2, and 3.