Question

Answer The Question In the picture Please

Answer 1

Answer:

4.(i)f(x)=2$\sqrt{2}$$x^{2}$-5$\sqrt{x}$-3$\sqrt{2}$,g(x)=x-2,

Now assuming g(x) is a factor and if f(x)=0, then g(x) is a factor,

x-2=0 [∵g(x) is a factor]

∴x=2

Putting x=2 in the equation,

f(x)=2$\sqrt{2}$×4-5

⇒f(x)=0

∴g(x) is a factor of the equation

(ii)f(x)=2$x^{4\\}$-12$x^{3}$+18x+14 ,g(x)=x-2,

Proceeding in this question as per above question,

f(x)=32-96+36+14=-14

∴g(x) is not a factor of f(x) because it is not equal to 0.

(iii)f(x)=7$x^{2}$-2$\sqrt{8}$x-6, g(x)=x-$\sqrt{2}$,

Proceeding as per above question,

∴x=$\sqrt{2}$ and put it in above equation.

f(x)=14-8-6=0

∴g(x) is a factor of f(x).

5.p(x)=$x^{3}$-3$x^{2}$-ax+b, factors  are (x+1) and (x+5),

∵(x+1) is a factor of p(x),

∴p(x)=-1-3+a+b[putting x=-1 as x+1 is a factor]

0=-4+a+b

a+b=4 ----equation 1

Now,

p(x)=-125-75+5a+b[putting x=-5 since x+5 is a factor]

200=5a+b -----equation 2

Subtracting equation 1 from equation 2,we get

5a+b-a-b=196

4a=196

⇒a=49

Now putting a=49 in equation 1,weget

⇒b= -45

∴a=49 and b= -45

Hope it helps.

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