Question

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→ If 2 and -2 are the zeros of polynomial p(x)=ax^4+2x^3-3x^2+bx-4, then value of a-b
→ If 2 and 1/2 are the zeros of polynomial p(x)=ax^2+5x+b, then value of a-b
→ If 1/2 is the zero of polynomial p(x)=2x^4-ax^3+4x^2+2x+1, then value of a
→ If 1 and -1 re the zeros of polynomial p(x)=ax^3+x^2-2x+b, find value of a+b

Answer 1

Given Polynomial : $\sf{p(x)=ax^4+2x^3-3x^2+bx-4}$

$\sf{p(2)\implies a(2)^4+2(2)^3-3(2)^2+b(2)-4=0}$

$\to\sf{p(2)\implies a(16)+2(8)-3(4)+b(2)-4=0}$

$\to\sf{p(2)\implies 16a+16-12+2b-4=0}$

$\to\sf{p(2)\implies 16a+4+2b-4=0}$

$\to\sf{p(2)\implies 16a+2b=0}$

$\to\sf{p(2)\implies 2(8a+b)=0}$

$\to\sf{p(2)\implies 8a+b=0\:\:\bf{.....(1)}}$

$\sf{p(-2)\implies a(-2)^4+2(-2)^3-3(-2)^2+b(-2)-4=0}$

$\to\sf{p(-2)\implies a(16)+2(-8)-3(4)+(-2b)-4=0}$

$\to\sf{p(-2)\implies 16a+(-16)-12-2b-4=0}$

$\to\sf{p(-2)\implies 16a-16-12-2b-4=0}$

$\to\sf{p(-2)\implies 16a-28-2b-4=0}$

$\to\sf{p(-2)\implies 16a-28-4-2b=0}$

$\to\sf{p(-2)\implies 16a-32-2b=0}$

$\to\sf{p(-2)\implies 2(8a-16-b)=0}$

$\to\sf{p(-2)\implies 8a-16-b=0\:\:\bf{.....(2)}}$

✧ Adding Equation 1 and 2

$\begin{array}{c}\sf{ \:\:\:8 a+b+\:0=0} \\\\\sf{+\:8 a+b-16=0} \\\frac {\qquad\qquad\qquad\qquad}{} \\\sf{\:\:\:16 a+2 b-16=0}\end{array}$

$\to\sf{16a-16=0}$

$\to\sf{16a-16+16=0+16}$

$\to\sf{16a=16}$

$\to\sf{a=1}$

✧ Keep This value in Equation 1

$\to\sf{8a+b=0}$

$\to\sf{8(1)+b=0}$

$\to\sf{8+b=0}$

$\to\sf{8+b-8=0-8}$

$\to\sf{b=-8}$

✧ We are Asked to Find Value of a - b

$\to\sf{a-b=1-(-8)}$

$\to\sf{a-b=1+8}$

$\large\boxed{\boxed{\to\sf{a-b=9}}}$

• Kindly Ask Question One By One

Answer 2

$\large\mathbb\pink{a - b = 9}$