If (x + 1) is a factor of x^4 + 9x^3 + 7x^2 + 9ax + 5a^2, then : -Maths 9th

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Description : If x-2 is a factor of x3-3x+5a then find the value of a -Maths 9th

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Description : 3. Check whether 7+3x is a factor of 3x3+7x. -Maths 9th

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Last Answer : (c) 4log3 (3 + x) + log3 (8 - x) - log3 (9x - 8) = 2 - log39 ⇒ log3 (3 + x) + log3 (8 - x) - log3 (9x - 8) + log39 = 2⇒ log3 \(\bigg[rac{(3+x)(8-x)(9)}{(9x-8)}\bigg]=2\)⇒ \(rac{9(24+8x-3x-x^2)}{(9x- ... = 9x - 8 ⇒ x2 + 4x - 32 = 0 ⇒ (x + 8) (x - 4) = 0 ⇒ x = - 8, 4. Taking the positive value x = 4.

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Description : Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. -Maths 9th

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Description : Find the value of the polynomial 3x3 – 4x2 + 7x – 5, when x = 3 and also when x = -3. -Maths 9th

Last Answer : Let p(x) =3x3 – 4x2 + 7x – 5 At x= 3, p(3) = 3(3)3 – 4(3)2 + 7(3) – 5 = 3×27-4×9 + 21-5 = 81-36+21-5 P( 3) =61 At x = -3, p(-3)= 3(-3)3 – 4(-3)2 + 7(-3)- 5 = 3(-27)-4×9-21-5 = -81-36-21-5 = -143 p(-3) = -143 Hence, the value of the given polynomial at x = 3 and x = -3 are 61 and -143, respectively.

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Description : Factorise : x2 + 9x +18 -Maths 9th

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Description : If 9x^2 + 3px + 6q when divided by (3x + 1) leaves a remainder (-3/4) -Maths 9th

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Description : One of the zeroes of the polynomial 2x2 + 7x – 4 is -Maths 9th

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Description : One of the zeroes of the polynomial 2x2 + 7x – 4 is -Maths 9th

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Description : If x +1 is a factor of ax3 +x2 -2x + 4a - 9, then find the value of a. -Maths 9th

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Description : If x + 2a is a factor of a5 -4a2x3 +2x + 2a +3, then find the value of a. -Maths 9th

Last Answer : Let p(x) = a5 -4a2x3 +2x + 2a +3 Since, x + 2a is a factor of p(x), then put p(-2a) = 0 (-2a)5 – 4a2 (-2a)3 + 2(-2a) + 2a + 3 = 0 ⇒ -32a5 + 32a5 -4a + 2a+ 3 = 0 ⇒ -2a + 3 = 0 2a =3 a = 3/2. Hence, the value of a is 3/2.

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Description : Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Last Answer : Solution: p(x) = 2x3+x2–2x–1, g(x) = x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1 ∴Zero of g(x) is -1. Now, p(−1) = 2(−1)3+(−1)2–2(−1)–1 = −2+1+2−1 = 0 ∴By factor theorem, g(x) is a factor of p(x

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Description : x + 1 is a factor of the polynomial -Maths 9th

Last Answer : (b) Let assume (x + 1) is a factor of x3 + x2 + x+1. So, x = -1 is zero of x3 + x2 + x+1 (-1)3 + (-1)2 + (-1) + 1 = 0 ⇒ -1+1-1 + 1 = 0 ⇒ 0 = 0 Hence, our assumption is true.

Description : Which of the following is a factor of (x+ y)3 – (x3 + y3) ? -Maths 9th

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Description : Determine which of the following polynomial has x – 2 a factor -Maths 9th

Last Answer : first option is the correct answer for the given question solution is as follows:- let x-2=0 then, x=2 put x in (i) 3(2)(2)+6(2)-24=0 12+12-24=0 {use BODMAS rule for solution}... 24-24=0 0=0 this verifies our answer