Question

If polynomial p (x) = x^3 + ax^2 + bx - 84 is exactly divisible by x^2 + x-12, find the values of a and b.​

Answer 1

Step-by-step explanation:

Given: If $p(x) = {x}^{3} + a {x}^{2} + bx - 84$is completely divisible by ${x}^{2} + x - 12$.

To find: Find the value of a and b.

Solution:Divide p(x) by $x^2+x-12$ and equate remainder to zero.

Alternative Method:

Step 1: Find zeros of $x^2+x-12$.

${x}^{2} + 4x - 3x - 12 = 0 \\ \\ x(x + 4) - 3(x + 4) = 0 \\ \\ (x + 4)(x - 3) = 0 \\ \\ (x + 4) = 0 \\ \\\bold{\green{ x = - 4} }\\ \\ or \\ \\ x - 3 = 0 \\ \\ \bold{\pink{x = 3}}$

Step 2: Put the zeros of $x^2+x-12$ in p(x).

If p(x) is completely divisible then according to remainder theorem; remainder will be zero.

$p( - 4) = 0 \\ {( - 4)}^{3} + a {( - 4)}^{2} + b( - 4) - 84 = 0 \\ \\ - 64 + 16a - 4b - 84 = 0 \\ \\ 16a - 4b - 148 = 0 \\ \\ or \\ \\ \bold{\red{4a - b - 37 = 0}}...eq1$

and

$p(3) = 0 \\ \\ {(3)}^{3} + a {(3)}^{2} + b(3) - 84 = 0 \\ \\ 27 + 9a + 3b - 84 = 0 \\ \\ 9a + 3b - 57 = 0 \\ \\ or \\ \\ \bold{\purple{3a + b - 19 = 0}}...eq2$

Step 3: Solve eq1 and eq2 to find a and b

Add both equations

$4a - b = 37 \\ 3a + b = 19 \\ - - - - - - \\ 7a = 56 \\ \\ a = \frac{56}{7} \\ \\ a = 8$

put the value of a in eq1

$4(8) - b = 37 \\ \\32 - b = 37 \\ \\ - b = 37 - 32 \\ \\ - b = 5 \\ \\ b = - 5$

Final answer:

$\bold{\red{a = 8}} \\ \\ \bold{\green{b = -5 }}$

Hope it helps you.

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2) If one of the Zeroes of the quadratic polynomials (a-1)x+ ax+1= -3, then find the value of a. https://brainly.in/question/41118278