1. Find the value of the polynomial ` \5x-4x^2+3\ ` at
(i) ` \x=0\ `
(ii) ` \x=-1\ `
(iii) ` \x=2\ `
solution :
(i) ` \x=0\ `
Let,
` \p(x)=5x-4x^2+3\ `
The value of the polynomial ` \p(x)\ ` at ` \x=0\ ` is given
` \p(0)=5×(0)-4×(0)^2+3\ `
` \=0-0+3\ `
` \=3\ `
(ii) ` \x=-1\ `
Let,
` \p(x)=5x-4x^2+3\ `
The value of the polynomial ` \p(x)\ ` at ` \x=-1\ ` is given
` \p(-1)=5×(-1)-4×(-1)^2+3\ `
` \=(-5)-4+3\ `
` \=-5-4+3\ `
` \=-9+3\ `
` \=-6\ `
(iii) ` \x=2\ `
Let,
` \p(x)=5x-4x^2+3\ `
The value of the polynomial ` \p(x)\ ` at ` \x=2\ ` is given
` \p(2)=5×(2)-4×(2)^2+3\ `
` \=10-(4×4)+3\ `
` \=10-16+3\ `
` \=-6+3\ `
` \=-3\ `
2. Find ` \p(0)\ `, ` \p(1)\ ` and ` \p(2)\ ` for each of the following polynomials:
(i) ` \p(y)=y^2-y=1\ `
(ii) ` \p(t)=2+t+2t^2-t^3\ `
(iii) ` \p(x)=x^3\ `
(iv) ` \p(x)=(x-1)(x+1)\ `
solution :
(i) ` \p(y)=y^2-y=1\ `
` \p(y)=y^2-y=1\ `
When, ` \p(0)\ `
` \p(0)=0^2-0+1\ `
` \=0-0+1\ `
` \=1\ `
When, ` \p(1)\ `
` \p(1)=1^2-1+1\ `
= 1 − 1 + 1
` \=0+1\ `
` \=1\ `
When, ` \p(2)\ `
` \p(2)=2^2-2+1\ `
` \=4-2+1\ `
` \2+1\ `
` \=3\ `
(ii) ` \p(t)=2+t+2t^2−t^3\ `
` \p(t)=2+t+2t^2−t^3\ `When, ` \p(0)\ `
` \p(0)=2+ 0+2×(0)^2−0^3\ `
` \=2+0−0\ `
` \=2\ `
When, ` \p(1)\ `
` \p(1)= 2 + 1 + 2×(1)^2−1^3\ `
` \=3+2-1\ `
` \=5-1\ `
` \=4\ `
When, ` \p(2)\ `
` \p(2)=2+2+2×(2)^2-2^3\ `
` \=4+8-8\ `
` \=4+0\ `
` \=4\ `
(iii) ` \p ( x ) = x^3\ `
` \p ( x ) = x^3\ `When, ` \p ( 0 )\ `
` \P(0)=0^3\ `
` \= 0\ `
When, ` \p ( 1 )\ `
` \p(1)=1^3\ `
` \= 1\ `
When, ` \P(2)\ `
` \p(2)=2^3\ `
` \= 8\ `
(iv) ` \P(x)=(x−1)(x+1)\ `
P ( x ) = ( x − 1 ) ( x + 1 )
When, ` \p(0)\ `
` \p(0)=(0-1)(0+1)\ `
` \=-1×1\ `
` \= −1\ `
When, ` \p ( 1 )\ `
` \p(1)=(1-1)(1+1)\ `
` \= 0 × 2\ `
` \= 0\ `
When, ` \p ( 2 )\ `
` \p(2)=(2-1)(2+1)\ `
` \= 1 × 3\ `
` \= 3\ `
3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) ` \p(x)=3x+1, x=\frac{-1}{3}\ `
(ii) ` \p(x)=5x-\pi, x=\frac{4}{5}\ `
(iii) ` \p(x)=x^2-1,x=1,-1\ `
(iv) ` \p(x)=(x+1)(x-2)\ `, ` \x=-1,2\ `
(v) ` \p(x)=x^2,x=0\ `
(vi) ` \p(x)=lx+m\ `, ` \x=\frac{-m}{l}\ `
(vii) ` \p(x)=3x^2-1\ `, ` \x=\frac{-1}{\sqrt{3}},\frac{2}{\sqrt{3}}\ `
(viii) ` \p(x)=2x+1,x=\frac{1}{2}\ `
solution :
(i) ` \p(x)=3x+1, x=\frac{-1}{3}\ `
For, `\x=\frac{-1}{3}\ `,
` \p(x)=3x+1\ `
` \p(\frac{-1}{3})=3(\frac{-1}{3})+1\ `
` \= −1 + 1\ `
` \= 0\ `
So, ` \frac{-1}{3}\ ` is a zero of ` \p(x)\ `.
(ii) ` \p(x)=5x-\pi, x=\frac{4}{5}\ `
For, ` \x=\frac{4}{5}\ `,
` \p(x)=5x-\pi\ `
` \p(\frac{4}{5})=5×(\frac{4}{5})-\pi\ `
` \=4−π\ `
So, ` \frac{4}{5}\ ` is not a zero of ` \p(x)\ `.
(iii) ` \p(x)=x^2-1,x=1,-1\ `
For, ` \x=1\ `,
` \p(x)=x^2-1,\ `
` \p(1)=1^2-1\ `
` \=1−1\ `
` \=0\ `
For, ` \x=-1\ `,
` \p(x)=x^2-1\ `,
` \p(-1)=-1^2-1\ `
` \=1−1\ `
` \=0\ `
So, 1 and -1 is a zero of ` \p(x)\ `.
(iv) ` \p(x)=(x+1)(x-2)\ `, ` \x=-1,2\ `
For, ` \x=-1\ `,
` \p(x)=(x+1)(x-2),\ `
` \p(-1)=(-1+1)(-1– 2)\ `
` \=0×(-3)\ `
` \= 0\ `
For, ` \x=2,\ `
` \p(x)=(x+1)(x-2)\ `,
` \p(2)=(2+1)(2–2)\ `
` \=3×0\ `
` \= 0\ `
So, -1 and 2 is a zero of ` \p(x)\ `.
(v) ` \p(x)=x^2,x=0\ `
For, ` \x=0\ `,
` \p(x)=x^2\ `
` \p(0)=0^2\ `
` \=0\ `
So, 0 is a zero of ` \p(x)\ `.
(vi) ` \p(x)=lx+m,x=\frac{-m}{l}\ `
For, ` \x=\frac{-m}{l}\ `
` \p(x)=lx+m\ `
` \p(\frac{-m}{l})=l(\frac{-m}{l})+m\ `
` \=-m+m\ `
` \=0\ `
So, ` \frac{-m}{l}\ ` is a zero of ` \p(x)\ `.
(vii) ` \p(x)=3x^2-1` , ` \x=\frac{-1}{\sqrt{3}}\ `,` \frac{2}{\sqrt{3}}\ `
For, ` \x=\frac{-1}{\sqrt{3}\ `
` \p(x)=3x^2-1\ `
` \p(\frac{-1}{\sqrt3})=3(\frac{-1}{\sqrt3})^2-1\ `
` \=3×(\frac{1}{3})-1\ `
` \=1-1\ `
` \=0\ `
For, ` \x=\frac{2}{\sqrt{3}\ `
` \p(x)=3x^2-1\ `
` \p(\frac{2}{\sqrt{3}})=3(\frac{2}{\sqrt{3}})^2-1\ `
` \=3×(\frac{4}{3})-1\ `
` \=4-1\ `
` \=3\ `
` \=3 ≠ 0\ `
So, ` \frac{-1}{\sqrt{3})\ ` is a zero of ` \p(x)\ ` but ` \(\frac{2}{\sqrt{3}})\ ` is not a zero of ` \p(x)\ `.
(viii) ` \p(x)=2x+1,x=\frac{1}{2}\ `
For, ` \x=\frac{1}{2}\ `,
` \p(x)=2x+1\ `
` \p(\frac{1}{2})=2(\frac{1}{2})+1\ `
` \=1+1\ `
` \=2\ `
` \=2≠0\ `
So, ` \frac{1}{2}\ ` is not a zero of `\p(x)\ `.
4. Find the zero of the polynomial in each of the following cases:
(i) ` \p(x)=x+5\ `
(ii) ` \p(x)=x–5\ `
(iii) ` \p(x)=2x+5\ `
(iv) ` \p(x)=3x-2\ `
(v) ` \p(x)=3x\ `
(vi) ` \p(x)=ax\ `, ` \a≠0\ `
(vii) ` \p(x)=cx+d\ `, ` \c≠0, c, d\ ` are real numbers.
solution :
(i) ` \p(x)=x+5\ `
` \⇒x+5=0\ `
` \⇒x=−5\ `
So, -5 is a zero of the polynomial ` \p(x)\ `.
(ii) ` \p(x)=x–5\ `
` \p(x)=x–5\ `
` \⇒x-5=0\ `
` \⇒x=5\ `
So, 5 is a zero of the polynomial ` \p(x)\ `.
(iii) ` \p(x)=2x+5\ `
` \⇒2x+5=0\ `
` \⇒2x=-5\ `
` \⇒x=\frac{-5}{2}\ `
So, ` \frac{-5}{2}\ ` is a zero of the polynomial ` \p(x)\ `.
(iv) ` \p(x)=3x-2\ `
` \⇒3x-2=0\ `
` \⇒3x=2\ `
` \⇒x=\frac{2}{3}\ `
So, ` \frac{2}{3}\ ` is a zero of the polynomial p(x).
(v) ` \p(x)=3x\ `
` \⇒3x=0\ `
` \⇒x=\frac{0}{3}\ `
` \⇒x=0\ `
So, 0 is a zero of the polynomial ` \p(x)\ `.
(vi) ` \p(x)=ax\ `, ` \a≠0\ `
` \⇒ax=0\ `
` \⇒x=\frac{0}{a}\ `
` \⇒x=0\ `
So, 0 is a zero of the polynomial ` \p(x)\ `.
(vii) ` \p(x)=cx+d\ `, ` \c≠0, c, d\ ` are real numbers.
` \⇒cx+d=0\ `
` \⇒cx=−d\ `
` \⇒x=\frac{-d}{c}\ `
So, ` \frac{-d}{c}\ ` is a zero of the polynomial ` \p(x)\ `.
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