Question

Find the value of the polynomial 5x – 4x2 + 3 at

(i) x = 0
(ii) x = – 1
(iii) x = 2

Give full explanation!​

Answer 1

Answer:

Given that:

(a + b + c)² = 32

ab + bc + ca = 10

To find:

a² + b² + c²

Solution:

ATQ:

\sf \Longrightarrow (a + b + c)^2 = 32⟹(a+b+c)

2

=32

| We know that:

| (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

\sf \Longrightarrow a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = 32⟹a

2

+b

2

+c

2

+2ab+2bc+2ac=32

Taking 2 as a common term from 2ab, 2bc & 2ac we get:

\sf \Longrightarrow a^2 + b^2 + c^2 + 2(ab + bc + ac) = 32⟹a

2

+b

2

+c

2

+2(ab+bc+ac)=32

| ATQ:

| ab + bc + ca = 10

| Substitute this value above.

\sf \Longrightarrow a^2 + b^2 + c^2 + 2(10) = 32⟹a

2

+b

2

+c

2

+2(10)=32

\sf \Longrightarrow a^2 + b^2 + c^2 + 20 = 32⟹a

2

+b

2

+c

2

+20=32

\sf \Longrightarrow a^2 + b^2 + c^2 = 32 - 20⟹a

2

+b

2

+c

2

=32−20

\sf \Longrightarrow a^2 + b^2 + c^2 = 12⟹a

2

+b

2

+c

2

=12

Final answer: 12

Answer 2

$\huge \sf {\orange {\underline {\pink{\underline{Answer :-}}}}}$

Let$\rm{ p(x) = 5x – 4x2 + 3}$

_________________________

(i) $\rm{p(0) = 5(0) – 4(0)2 + 3 }$

$\rm{= 0 – 0 + 3 = 3}$

Thus, the value of

$\rm{5x – 4x² + 3 at x = 0~is ~3.}$

_________________________

(ii) $\rm{p(-1) = 5(-1) – 4(-1)2 + 3}$

$\rm{= – 5x – 4x2 + 3 }$

$\rm{= -9 + 3 = -6}$

Thus, the value of

$\rm{5x – 4x2 + 3 ~at~ x -1~ is~ -6.}$

_________________________

(iii) $\rm{p(2) = 5(2) – 4(2)2 + 3 }$

$\rm{= 10 – 4(4) + 3}$

$\rm{= 10 – 16 + 3 = -3}$

Thus, the value of

$\rm{5x – 4x2 + 3 at x = 2 is – 3.}$

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