Question

Find the value of each of the following polynomial:​

Answer 1

Answer:

Step-by-step explanation:

( i )

$p(x) = 4x^2 -3x + 7\\\\p(1) = 4(1)^2 - 3(1) + 7 = 4 - 3 + 7 = 1 + 7 = 8$

( ii)

$q(y) = 2y ^3 -4y + \sqrt{11}\\\\q(1) = 2( 1)^3 - 4(1) + \sqrt{11} = 2 - 4 + \sqrt{11} = -2 + \sqrt{11}$

(iii)

$r(t) = 4t^4 + 3t^3 -t^2+6\\\\r(p) = 4(p)^4 + 3(p)^3 - (p)^2 + 6= 4p^4 + 3p^3 -p^2 + 6$

(iv)

$s(z) = z^3 -1\\s(1) = (1)^3 -1 = 1 - 1 = 0$

(v)

$p(x) = 3x^2+5x-7 \\\\p(1) = 3(1)^2 + 5(1) - 7 = 3 + 5 - 7 = 8 - 7 = 1$

(vi)

$q(z) =5z^3 -4z +\sqrt{2}\\\\q(2) = 5(2)^3 - 4(2) \sqrt2 = (5 \times 8) - ( 4 \times 2) + \sqrt 2 = 40 - 8 + \sqrt 2 = 32 + \sqrt2$

Answer 2

Step-by-step explanation:

( i )

\begin{gathered}p(x) = 4x^2 -3x + 7\\\\p(1) = 4(1)^2 - 3(1) + 7 = 4 - 3 + 7 = 1 + 7 = 8\end{gathered}

p(x)=4x

2

−3x+7

p(1)=4(1)

2

−3(1)+7=4−3+7=1+7=8

( ii)

\begin{gathered}q(y) = 2y ^3 -4y + \sqrt{11}\\\\q(1) = 2( 1)^3 - 4(1) + \sqrt{11} = 2 - 4 + \sqrt{11} = -2 + \sqrt{11}\end{gathered}

q(y)=2y

3

−4y+

11

q(1)=2(1)

3

−4(1)+

11

=2−4+

11

=−2+

11

(iii)

\begin{gathered}r(t) = 4t^4 + 3t^3 -t^2+6\\\\r(p) = 4(p)^4 + 3(p)^3 - (p)^2 + 6= 4p^4 + 3p^3 -p^2 + 6\end{gathered}

r(t)=4t

4

+3t

3

−t

2

+6

r(p)=4(p)

4

+3(p)

3

−(p)

2

+6=4p

4

+3p

3

−p

2

+6

(iv)

\begin{gathered}s(z) = z^3 -1\\s(1) = (1)^3 -1 = 1 - 1 = 0\end{gathered}

s(z)=z

3

−1

s(1)=(1)

3

−1=1−1=0

(v)

\begin{gathered}p(x) = 3x^2+5x-7 \\\\p(1) = 3(1)^2 + 5(1) - 7 = 3 + 5 - 7 = 8 - 7 = 1\end{gathered}

p(x)=3x

2

+5x−7

p(1)=3(1)

2

+5(1)−7=3+5−7=8−7=1

(vi)

\begin{gathered}q(z) =5z^3 -4z +\sqrt{2}\\\\q(2) = 5(2)^3 - 4(2) \sqrt2 = (5 \times 8) - ( 4 \times 2) + \sqrt 2 = 40 - 8 + \sqrt 2 = 32 + \sqrt2\end{gathered}

q(z)=5z

3

−4z+

2

q(2)=5(2)

3

−4(2)

2

=(5×8)−(4×2)+

2

=40−8+

2

=32+

2