Question

Find the value of a and b so that 1, -2 are the zeroes of the polynomial x^3 + 10x^2 + ax + b??

I didn't understand how to solve this problem, thanks in advance...

Answer 1

Answer:-

Given:

• 1 and -2 are zeroes of given eq.
• ${x}^{3} + 10 {x}^{2} + ax + b$

Find:

• a and b =?

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$\huge{\mathbb{\underline{\red{SOLUTION}}}}$

Let the equation be p(x)

As 1 is a solution of p(x),

then p(1) must be equal to 0.

putting 1 in the place of x in p(x),

${x}^{3} + 10 {x}^{2} + a x + b \\ \implies \: ( {1}^{3} ) + 10 \times ( {1}^{2} ) + 1 \times a + b = 0 \\ \implies \: 1 + 10 + a + b = 0 \\ \implies \: a + b = ( - 11)-------i$

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Now also given,

(-2) is a zero of the polynomial,

thus p(-2)=0

Putting x=(-2), in p(x) we get,

${x}^{3} + 10 {x}^{2} + a x + b \\ \implies \: ( { - 2}^{3} ) + 10 \times ( { - 2}^{2} ) + ( - 2)\times a + b = 0 \\ \implies \: ( - 8) + 4 \times 10 - 2a + b = 0 \\ \implies \: ( - 8) + 40 - 2a + b = 0 \\ \implies \: 32 - 2a + b = 0 \\ \implies \: 32 = 2a - b------ii$

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Adding i and ii we get,

$2a - b + a + b = 32 - 11 \\ \implies \: 3a = 21 \\ \implies \: a = \frac{21}{3}$

Thus $\large{\red{\boxed{\boxed{a=7}}}}$

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Now we know from i,

a+b=(-11)

=> 7+b =(-11)

=> b= (-11-7)

=> b=(-18)

$\large{\red{\boxed{\boxed{b=(-18)}}}}$