<span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">Show that ( x + 2 ) is a factor of f ( x ) = x 3 + 4 x 2 + x − 6 .

By the factor theorem, ( x + 2 ) will be a factor of f ( x ) if f ( − 2 ) = 0 . Now, f ( x ) = x 3 + 4 x 2 + x − 6 ⇒ f ( − 2 ) = ( − 2 ) 3 + 4 ( − 2 ) 2 + ( − 2 ) − 6 = − 8 + 16 − 2 − 6 = 0 Hence, ( x + 2 ) is factor of f ( x ) = x 3 + 4 x 2 + x − 6 .

<span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">Show that the polynomial f ( x ) = x 4 + 4 x 2 + 6 <span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;"> has no zero.

f ( x ) = x 4 + 4 x 2 + 6 = ( x 2 ) 2 + 4 x 2 + 6 Put x 2 = a Then, f ( x ) = a 2 + 4 a + 6 , Now we need to check whether the f ( x ) has real roots or not, find the discriminant. D = b 2 - 4 a c = 16 - 24 = - 8 < 0 We know that the zeroes of the polynomial are the real values of x for which D ≥ 0 . Since we are not getting any real roots for f ( x ) , so there is no zero for f ( x ) . Hence, f ( x ) = x 4 + 4 x 2 + 6 has no zero.

<span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">If one zero of the polynomial p ( x ) = x 3 − 6 x 2 + 11 x − 6 <span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;"> is 3 , find the other two zeros.

Let α , β , γ are the zeros of the polynomial p ( x ) = x 3 − 6 x 2 + 11 x − 6 , we have α + β + γ = - − 6 1 = 6 and α β γ = - - 6 1 = 6 Substituting α = 3 , ⇒ β + γ = 3 and β γ = 2 So, β and γ can be 2 and 1 Hence, the other zeros of the polynomial are 1 & 2

If two zeros of the polynomial p ( x ) = 2 x 4 − 3 x 3 − 3 x 2 + 6 x − 2 are 2 and - 2 , find its other two zeroes.

The given polynomial is p ( x ) = 2 x 4 − 3 x 3 - 3 x 2 + 6 x − 2 . Since 2 and - 2 are the zeros of p ( x ) , it follows that each one ( x − 2 ) and ( x + 2 ) is a factor of p ( x ) . Consequently, ( x − 2 ) ( x + 2 ) = ( x 2 − 2 ) is factor of p ( x ) . On dividing p ( x ) by ( x 2 − 2 ) , we get ∴ p ( x ) = 0 ⇒ ( x 2 − 2 ) ( 2 x 2 − 3 x + 1 ) = 0 ⇒ ( x 2 − 2 ) ( 2 x 2 − 2 x - x + 1 ) = 0 ⇒ ( x − 2 ) ( x + 2 ) 2 x ( x − 1 ) − 1 ( x - 1 ) = 0 ⇒ ( x − 2 ) ( x + 2 ) ( x − 1 ) ( 2 x − 1 ) = 0 ⇒ ( x − 2 ) = 0 or ( x + 2 ) = 0 or x - 1 = 0 or ( 2 x − 1 ) = 0 ⇒ x = 2 or ⇒ x = - 2 or x = 1 or x = 1 2 Thus, the other two zeros are 1 and 1 2 .

<span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">Find the quotient when p ( x ) = 3 x 4 + 5 x 3 − 7 x 2 + 2 x + 2 <span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;"> is divided by ( x 2 + 3 x + 1 ) <span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">.

By long division, ∴ Quotient = 3 x 2 - 4 x + 2

<span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">Use remainder theorem to find the value of k , it being given that when x 3 + 2 x 2 + k x + 3 <span style="color: rgb(0, 0, 0); font-family: Roboto, sans-serif; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;"> is divided by ( x - 3 ) , then the remainder is 21 .

Let p ( x ) = x 3 + 2 x 2 + k x + 3 By the remainder theorem, we know that when p ( x ) is divided by ( x − 3 ) the remainder is p ( 3 ) . Now, p ( 3 ) = ( 3 ) 3 + 2 ( 3 ) 2 + k ( 3 ) + 3 = 27 + 18 + 3 k + 3 = 48 + 3 k But, remainder = 21 ⇒ 48 + 3 k = 21 ⇒ 3 k = − 27 ⇒ k = − 9

E M B I B E

Secondary School Mathematics for Class 10>Algebra>Chapter 2 - Polynomials>TEST YOURSELF

R S Aggarwal and Veena Aggarwal Solutions for Exercise 6: TEST YOURSELF

Author:R S Aggarwal & Veena Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal and Veena Aggarwal Solutions for Exercise 6: TEST YOURSELF

Attempt the free practice questions from Exercise 6: TEST YOURSELF with hints and solutions to strengthen your understanding. Secondary School Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from R S Aggarwal and Veena Aggarwal Solutions for Exercise 6: TEST YOURSELF with Hints & Solutions

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>TEST YOURSELF>Q 13

Show that $(x + 2)$ is a factor of $f (x) = x^{3} + 4 x^{2} + x - 6$ .

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>TEST YOURSELF>Q 14

If $α, β, γ$ are the zeros of the polynomial $p (x) = 6 x^{3} + 3 x^{2} - 5 x + 1$ , find the value of $(\frac{1}{α} + \frac{1}{β} + \frac{1}{γ})$ .

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>TEST YOURSELF>Q 15

If $α, β$ are the zeros of the polynomial $f (x) = x^{2} - 5 x + k$ such that $α - β = 1$ , find the value of $k$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>TEST YOURSELF>Q 16

Show that the polynomial $f (x) = x^{4} + 4 x^{2} + 6$ has no zero.

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>TEST YOURSELF>Q 17

If one zero of the polynomial $p (x) = x^{3} - 6 x^{2} + 11 x - 6$ is $3$ , find the other two zeros.

HARD

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>TEST YOURSELF>Q 18

If two zeros of the polynomial $p (x) = 2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2$ are $\sqrt{2}$ and $- \sqrt{2}$ , find its other two zeroes.

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>TEST YOURSELF>Q 19

Find the quotient when $p (x) = 3 x^{4} + 5 x^{3} - 7 x^{2} + 2 x + 2$ is divided by $(x^{2} + 3 x + 1)$ .

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 2 - Polynomials>TEST YOURSELF>Q 20

Use remainder theorem to find the value of $k$ , it being given that when $x^{3} + 2 x^{2} + k x + 3$ is divided by $(x - 3)$ , then the remainder is $21$ .

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