For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th

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Description : If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Last Answer : Given that the following polynomials leave the same remainder when divided by (x - 4) : We are to find the value of a. Remainder theorem: When (x - b) divides a polynomial p(x), then the remainder is p(b). So, from (i) and (ii), we get Thus, the required value of a is 1.

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Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Last Answer : The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10

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Last Answer : Let α be the common root ∴α2+pα+q=0 ...........(1) and α2+qα+p=0 ........ (2) Solving (1) & (2), we get, p2−q2α2​=q−pα​=q−p1​∴α=q−pp2−q2​ and α=1 ⇒q−pp2−q2​=1 ⇒p2−q2=q−p (or) (p2−q2)+(p−q)=0 ⇒(p−q)[p+q+1]=0 ⇒p−q=0 or p+q+1=0

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Description : If x^3 + 5x^2 + 10k leaves remainder – 2x when divided by x^2 + 2, then what is the value of k ? -Maths 9th

Last Answer : x3+5x2+10k =(x2+2)(x+5)+10k−2x−10 ⇒10k−2x−10=−2x ⇒10k−10=0 or k=1.

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Last Answer : (a) Since, (2, 0) is a solution of the given linear equation 2x + 3y = k, then put x =2 and y= 0 in the equation. ⇒ 2 (2) + 3 (0) = k ⇒ k = 4 Hence, the value of k is 4.

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Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

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Description : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

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Description : Zero of the polynomial p(x)=2x+5 is -Maths 9th

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