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2021-12-10

Find p(0) , p(1) for the given polynomial

$p\left(x\right)=10x-4{x}^{2}-3$

Cleveland Walters

Beginner2021-12-11Added 40 answers

Step 1 Given

$p\left(x\right)=10x-4{x}^{2}-3$

Step 2 Finding

When x=0,

p(0)=10(0)-4(0)2-3

=0-0-3

=-3

when x=1

$p\left(x\right)=10x-4{x}^{2}-3$

$p\left(1\right)=10\left(1\right)-4{\left(1\right)}^{2}-3$

=10-4-3

=3

$\therefore p\left(x\right)=10x-4{x}^{2}-3=-3$ when x=0

$p\left(x\right)=10x-4{x}^{2}-3=3$ when x=1

Step 2 Finding

When x=0,

p(0)=10(0)-4(0)2-3

=0-0-3

=-3

when x=1

=10-4-3

=3

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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