Text Solution Solution : Given `p(x)=2x^4+7x^3−19x^2−14x+30`<br> Two of its zeroes are `sqrt(2) `and `−sqrt(2)`<br> so,`g(x)=(x−sqrt(2))(x+sqrt(2))`<br> or,`x^2−2`<br> so we get <br>,divisor=`x^2−2`<br> Quotient=`2x^2+7x−15`<br> Remainder=`0`<br> so,`q(x)=−2x^2+7x−15`<br> `2x^2+10x−3x−15=0`<br> =`2x(x+5)−3(x+5)=0`<br> =`(2x−3)(x+5)=0`<br> thus,`x=3/2,-5`<br>so,roots of equation is `sqrt(2) ,−sqrt(2) ,3/2 ` and `−5` No worries! We‘ve got your back. Try BYJU‘S free classes today! Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App Suggest Corrections 4
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and coefficients in each case: Let p(x) = 2x3+x2–5x + 2 Comparing the given polynomial with ax3 + bx3 + cx + d, we get Now,
and ∴ are the zeroes of Hence, verified.Here, we have Now, and Hence verified. |